representable if and only if $G$ is cyclic.

###### Proof.

One direction is clear, so we assume $G$ is freely representable. So by the
above theorem, the order of the kernel $K$ of the action homomorphism
$B\to\mathrm{Aut}(A)$ must be divisible by exactly the primes that divide the
order of $B$. This means that $K=B$, so $G=AB=A\times B$ is cyclic. ∎

We can restate the above lemma in terms $\mu(G)$:

######  Corollary 77.

Let $G$ be a freely representable Sylow-cyclic group, and let $\mu(G)$ be the
MCC subgroup of $G$. Then every element of $G$ of prime order is in $\mu(G)$.
In particular, every prime dividing the order of $G$ must divide the order of
$\mu(G)$.

###### Proof.

Write $G$ as $AB=A\rtimes B$ where $A$ and $B$ are cyclic subgroups of $G$ of
relatively prime orders. Let $g=ab$ be an element of prime order where $a\in
A$ and $b\in B$. Under the projection $A\rtimes B\to B$ the element $g$ maps
to $b$, so $b$ is of prime order or is the trivial element. In either case, by
the above lemma, $b\in K$ where $K$ is the kernel of the action map
$B\to\mathrm{Aut}(A)$. So $g=ab\in AK$. But $AK=\mu(G)$ (Lemma 71) so
$g\in\mu(G)$. ∎

This gives a necessary condition for a Sylow-cyclic group to be freely
representable. In order to show it is sufficient we use induced
representations, but in a very basic manner. The following, which we take as
given, is all that we need to know about induced representations here:

######  Proposition 78.

Let $G$ be a finite group with subgroup $H$, and let $F$ be a field. Suppose
that $H$ acts linearly on an $F$-vector space $W$. Then there is a linear
action of $G$ on an $F$-vector space $V$ containing $W$ such that (1) the
action of $G$ on $V$ restricts to the given action of $H$ on $W$, (2) if
$g_{1}H,\ldots,g_{k}H$ are the distinct left cosets in $G/H$ then the vector
space $W$ is the direct sum of the spaces $g_{i}W$:

$V=\bigoplus g_{i}W.$

This representation, called the induced representation, is unique up to a
$F[G]$-module isomorphism fixing $W$.

######  Proposition 79.

Let $G$ be a finite group with a subgroup $H$ that contains all elements of
$G$ of prime order. Suppose $W$ is an $F$-vector space with a linear
representation of $H$ on $W$, and suppose $V$ is a $F$-vector space containing
$W$ with a representation of $G$ induced by the representation of $H$ on $W$.
If the linear representation of $H$ on $W$ is a free linear representation,
then the linear representation of $G$ on $V$ is also a free linear
representation.

###### Proof.

Let $g_{1}H,\ldots,g_{k}H$ be the distinct left cosets in $G/H$. Suppose
$\sigma\in G$ is not the identity and that $\sigma(v)=v$ where $v\in V$ is
equal to

$v=g_{1}w_{1}+\ldots+g_{k}w_{k}$

where $w_{i}\in W$. Let $m>1$ be the order of $\sigma$ and let $p$ be a prime
dividing $m$. Then $\tau=\sigma^{m/p}$ has order $p$ and also fixes $v$. By
assumption $\tau\in H$. Observe that the conjugate $\tau_{i}=g_{i}^{-1}\tau
g_{i}$ also has order $p$ so is in $H$. Since $\tau g_{i}=g_{i}\tau_{i}$

$\tau v=\tau g_{1}w_{1}+\ldots+\tau
g_{k}w_{k}=g_{1}(\tau_{1}w_{1})+\ldots+g_{k}(\tau_{k}w_{k}).$

Since $\tau v=v$,

$g_{1}(\tau_{1}w_{1})+\ldots+g_{k}(\tau_{k}w_{k})=g_{1}w_{1}+\ldots+g_{k}w_{k}.$

By the direct sum property of induced representations,

$g_{i}(\tau_{i}w_{i})=g_{i}w_{i}$

for each $1\leq i\leq k$. Multiplying by the inverse of $g_{i}$ gives
$\tau_{i}w_{i}=w_{i}$ for the induced representation. Since $\tau_{i}\in H$
and $w_{i}\in W$, we have $\tau_{i}w_{i}=w_{i}$ in the original representation
of $H$ on $W$. This representation is a free linear representation by
assumption, so $w_{i}=0$ for each $i$. This implies $v=0$, showing that the
induced representation is a free linear representation. ∎

######  Corollary 80.

Let $G$ be a finite group with freely representable subgroup $H$. If $H$
contains all elements of $G$ of prime order, then $G$ is also freely
representable.

Now we are ready for the main theorem:

######  Theorem 81.

Let $G$ be a Sylow-cyclic group and let $\mu(G)$ be its maximal characteristic
cyclic subgroup. The following are equivalent:

  1. 1.

$G$ is freely representable.

  2. 2.

Every prime dividing the order of $G$ also divides the order of $\mu(G)$.

  3. 3.

For every prime $p$ dividing the order of $G$ there is a unique subgroup of
$G$ of order $p$.

###### Proof.

Suppose $G$ is freely representable. Then by Corollary 77 every prime dividing
the order of $G$ must divide the order of $\mu(G)$. So $(1)\implies(2)$.

Now assume $(2)$. If $p$ divides $|G|$ then, by assumption, $p$ divides
$|\mu(G)|$. Thus there is a unique subgroup of $G$ of order $p$ by Corollary
64. So $(2)\implies(3)$ holds.

Finally assume $(3)$. Let $g\in G$ be an element of prime $p$ order. By
assumption the subgroup $\left<g\right>$ is the unique subgroup of $G$ of
order $p$. This implies that $\left<g\right>$ is a characteristic cyclic
subgroup, and so $\left<g\right>$ is contained in $\mu(G)$. Since $\mu(G)$ is
cyclic, it is freely representable. Thus $G$ is freely representable by
Corollary 80. ∎

We can now strengthen Lemma 75.

######  Corollary 82.

Let $A$ and $B$ be cyclic subgroups of $G$ of relatively prime orders such
that $G=A\rtimes B$ (so $A$ is normal in $G$). Let $K$ be the kernel of the
action homomorphism $B\to{\mathrm{Aut}}(A)$ associated to the semidirect
product. Then $G$ is freely representable if and only if every prime dividing
$|B|$ divides $|K|$.

###### Proof.

Recall that $\mu(G)$ is $AK$ (Lemma 71).

If every prime dividing $|B|$ divides $|K|$ then every prime dividing the
order of $G$ must divide $|A|$ or $|K|$. Hence every prime dividing the order
of $G$ divides the order of $\mu(G)$, and so $G$ is freely representable by
the above theorem.

Conversely, if $G$ is freely representable, then every prime $p$ dividing the
order of $B$ must divide the order of $\mu(G)=AK\cong A\times K$ by the above
theorem. But $p$ does not divide the order of $A$, so $p$ divides the order of
$K$ as desired. ∎

######  Example 11.

Suppose $A$ a cyclic group of odd prime order $p$. Let $q$ be any prime
dividing $p-1$. Let $B$ be a cyclic group of order $q^{k}$ with $k>1$. We can
identify $\mathrm{Aut}(A)$ with ${\mathbb{F}}_{p}^{\times}$ which is cyclic of
order $p-1$. Let $B^{\prime}$ be the unique quotient of $B$ of size $q$, and
fix an injective homomorphism $B^{\prime}\to\mathrm{Aut}(A)$. Now have $B$ act
on $A$ by the composition

$B\to B^{\prime}\to\mathrm{Aut}(A)$

and let $G$ be the associated semidirect product $A\rtimes B$. The kernel $K$
of this action homomorphism has order $q^{k-1}$. By Lemma 71 we have that
$\mu(G)\cong A\times K$, and this has $pq^{k-1}$ elements. By the above
theorem, $G$ is a freely representable non-Abelian group of order $pq^{k}$. In
the case of order $p\cdot 2^{2}$ such groups arose already as binary dihedral
groups $2D_{p}$ in ${\mathbb{H}}^{\times}$. But if we take $q\neq 2$ we can
conclude the following: _There are an infinite number of odd orders such that
there exists non-Abelian Sylow-cyclic groups of that order that are freely
representable_ (for example, for a fixed $q$ take an infinite sequence of
primes $p\equiv 1\bmod q$). Note that such Sylow-cyclic groups cannot be
isomorphic to subgroups of ${\mathbb{H}}^{\times}$ since all finite non-
Abelian subgroups of ${\mathbb{H}}^{\times}$ have order divisible by $4$.

The smallest such order of this type of group of odd order is $7\cdot
3^{2}=63$. Note that if $G$ is a noncyclic freely representable group of order
$5\cdot 3^{2}$ then $G$ is $A\rtimes B$ where $A$ is cyclic subgroup of $G$ of
order $5$ and $B$ is a cyclic subgroup of $G$ of order $9$. Furthermore,
$\mu(G)$ has order $15$ or $45$. The second case cannot happen since $G$ is
not cyclic. The first case cannot happen either: the automorphism group of $A$
has order 4, so the kernel $K$ of the action of $B$ on $A$ must be all of $K$,
so $\mu(G)=AK=G$. We conclude that 63 is the smallest odd order possible for a
noncyclic freely representable group.

Here is another interesting application of Theorem 81.

######  Proposition 83.

Let $G$ be a freely representable Sylow-cyclic group and let $N$ be a normal
subgroup of $G$ of index $p$. If $p^{2}$ does not divide the order of $G$ then

$G=NC_{p}\cong N\times C_{p}$

where $C_{p}$ is a subgroup of $G$ of order $p$ (and is the unique subgroup of
order $p$).

###### Proof.

Since $G$ is freely representable, there is a unique subgroup $C_{p}$ of order
$p$, and so $C_{p}$ must be normal in $G$. Since $N\cap C_{p}=\\{1\\}$ we have
$NC_{p}\cong N\times C_{p}$. Finally, $G=NC_{p}$ since $[G:N]=p$. ∎

We cannot hope to generalize Theorem 81 to all freely representable groups.
For example, the binary tetrahedral group $2T$ is freely representable (as a
subgroup of ${\mathbb{H}}^{\times}$), but does not have a unique subgroup of
order 3. However, one implication holds in general:

######  Proposition 84.

Let $G$ be a finite group with the property that for each prime $p$ dividing
the order of $G$ there is a unique subgroup of order $p$. Then $G$ is freely
representable.

###### Proof.

Let $p_{1},\ldots,p_{k}$ be the primes dividing the order of $G$. Let
$C_{p_{i}}$ be the be the unique subgroup of order $p_{i}$. Then each
$C_{p_{i}}$ is normal and

$H\;\stackrel{{\scriptstyle\mathrm{def}}}{{\,=\,}}\;C_{p_{1}}\cdots
C_{p_{k}}\cong C_{p_{1}}\times\cdots\times C_{p_{k}}$

is a cyclic subgroup of $G$. So $H$ is freely representable. Now use Corollary
80. ∎

######  Remark.

The classification of Sylow-cyclic groups is enough to yield significant
applications to differential geometry. In fact, by a theorem of Vincent
(1947), every complete connected Riemannian manifold of constant positive
curvature of dimension not congruent to $3$ modulo $4$ has a fundamental group
that is Sylow-cyclic. From this Vincent was able to give a full classification
of such manifolds when the dimension is not congruent to $3$ modulo $4$. Wolf
[17] completed the classification to all dimensions by classifying freely
representable groups beyond the Sylow-cyclic groups.

###  2 Application to Automorphisms of Sylow-cyclic groups

Consider the automorphism group ${\mathrm{Aut}}(G)$ where $G$ is a Sylow-
cyclic group of odd order, and let $O({\mathrm{Aut}}(G))$ be the maximal odd
normal subgroup of ${\mathrm{Aut}}(G)$. Then we can use the above results to
show that ${\mathrm{Aut}}(G)/O({\mathrm{Aut}}(G))$ is an Abelian $2$-group.
This is clear if $G$ is cyclic since ${\mathrm{Aut}}(G)$ is an Abelian group.
The key to generalizing this is to relate this quotient to the corresponding
quotient for the cyclic subgroup $\mu(G)$. We start with a lemma.

######  Lemma 85.

Let $G$ be a Sylow-cyclic group of odd order and let $\phi$ be an automorphism
of $G$ such that $\phi^{2}$ is the identity map. If $\phi$ fixes the MCC
subgroup $\mu(G)$ then $\phi$ fixes all of $G$.

###### Proof.

Let $P$ be a nontrivial Sylow subgroup of $G$. We will show that $\phi$ acts
trivially on $P$. Since the Sylow subgroups of $G$ generate $G$, this gives
the result. Let $A$ be the commutator subgroup of $G$. Since $A$ and $G/A$
have relatively prime orders, either $P\subseteq A$ or $P\cap A=\\{1\\}$. In
the first case $\phi$ acts trivially on $P$ since $A\subseteq\mu(G)$. So from
now on we assume that $P\cap A=\\{1\\}$.

Observe that the image $\overline{P}$ of $P$ in $G/A$ is isomorphic to $P$.
Since $\overline{P}$ is a cyclic group of odd order, it has a unique
automorphism of order $2$. So $\phi$ acts on $\overline{P}$ either as
$x\mapsto x$ or as $x\mapsto x^{-1}$.

First suppose that $\phi$ acts on $\overline{P}$ as $x\mapsto x$. So if $b\in
P$ then $\phi(b)=ab$ for some $a\in A$. Observe then that

$b=\phi^{2}(b)=\phi(ab)=\phi(a)\phi(b)=a(ab)=a^{2}b.$

Thus $a^{2}=1$. Since $A$ is a cyclic group of odd order $a=1$, and
$\phi(b)=b$. We conclude that $\phi$ acts trivially on $P$.

Finally suppose that $\phi$ acts on $\overline{P}$ as $x\mapsto x^{-1}$ and
let $c\in P$ be a generator. Thus $\phi(c)=c^{-1}a_{0}$ for some $a_{0}\in A$.
Let $B$ be a complement of $A$ and observe that $P$ is conjugate to a Sylow
subgroup of $B$. So replacing $B$ with a complement of $B$ if necessary, we
can assume $P$ is a subgroup of $B$. Note that $\phi(c)\neq c$ since $c^{-1}$
and $c$ have distinct images in $\overline{P}$ (and $\overline{P}$ has odd
order greater than 1). Thus $c$ is not in the center $Z(G)$ since
$Z(G)\subseteq\mu(G)$. Since $c$ centralizes $B$, it cannot centralize $A$.
Let $a\in A$ be such that $cac^{-1}\neq a$. Note that $cac^{-1}\in A$ so

$cac^{-1}=\phi(cac^{-1})=\phi(c)a\phi(c)^{-1}=c^{-1}a_{0}aa_{0}^{-1}c=c^{-1}ac.$

So $c^{2}ac^{-2}=a$. However $c$ has odd order, so this implies that
$cac^{-1}=a$, a contradiction. So $\phi$ cannot act on $\overline{P}$ as
$x\mapsto x^{-1}$. ∎

######  Proposition 86.

Let $G$ be a Sylow-cyclic group of odd order. Let $\mu(G)$ be the MCC subgroup
of $G$, and let $O({\mathrm{Aut}}(G))$ be the maximal odd normal subgroup of
the automorphism group ${\mathrm{Aut}}(G)$. Then the quotient

${\mathrm{Aut}}(G)/O({\mathrm{Aut}}(G)).$

is isomorphic to a $2$-group inside ${\mathrm{Aut}}(\mu(G))$. In particular,
this quotient is an Abelian $2$-group.

###### Proof.

Observe that ${\mathrm{Aut}}(\mu(G))$ is Abelian since $\mu(G)$ is a cyclic
group. Recall that Abelian groups are the products of their Sylow subgroups,
and so

${\mathrm{Aut}}(\mu(G))=A_{1}A_{2}=A_{1}\times A_{2}$

where $A_{1}$ is the subgroup of ${\mathrm{Aut}}(\mu(G))$ consisting of
elements of odd order, and where $A_{2}$ is the $2$-Sylow subgroup of
${\mathrm{Aut}}(\mu(G))$. Since $\mu(G)$ is characteristic in $G$ we have a
homomorphism ${\mathrm{Aut}}(G)\to{\mathrm{Aut}}(\mu(G))$. We also have the
projection homomorphism ${\mathrm{Aut}}(\mu(G))=A_{1}A_{2}\to A_{2}$. Let $K$
be the kernel of the composition

${\mathrm{Aut}}(G)\to{\mathrm{Aut}}(\mu(G))\to A_{2}.$

Observe that $K$ contains $O({\mathrm{Aut}}(G))$.

Claim: $K$ contains only elements of odd order. Suppose otherwise that
$\psi\in K$ has order $2k$, and let $\phi=\psi^{k}$. Then $\phi$ has order $2$
and is in $K$. The image of $\phi$ in ${\mathrm{Aut}}(\mu(G))$ is just the
restriction $\phi|_{\mu(G)}$ and it is in the kernel of of the projection
${\mathrm{Aut}}(\mu(G))=A_{1}A_{2}\to A_{2}$. In other words,
$\phi|_{\mu(G)}\in A_{1}$ and so has odd order. Since $\phi$ has order $2$, we
conclude that $\phi|_{\mu(G)}$ has order 1. By the previous lemma $\phi$ is
the identity, a contradiction.

So the claim has been established. This means $K=O({\mathrm{Aut}}(G))$ and we
have an injection

${\mathrm{Aut}}(G)/O({\mathrm{Aut}}(G))\hookrightarrow A_{2}.$

The result follows. ∎

##  Chapter 7 Applications to Division Rings

The classification of Sylow-cyclic fields can be used to prove Wedderburn’s
theorem. Along the way we will see an argument that every finite subgroup of a
field is cyclic. This section is independent of Example 3 where we used (1)
Wedderburn’s theorem and (2) the fact that $F^{\times}$ is cyclic for any
finite field. We start with the following.

######  Lemma 87.

Let $D$ be a division ring and let $F$ be its prime subfield. Then every
finite subgroup $G$ of $D^{\times}$ is freely representable over $F$.

###### Proof.

We let $G$ act on $V=D$ by left multiplication. Note that $V$ is an $F$-vector
space. This action is a free linear action since $D$ has no zero divisors. ∎

######  Remark.

In particular, if $F$ has characteristic zero then $G$ is freely representable
(and so is Sylow-cycloidal). This result is the starting point for Amitsur’s
classification of finite subgroups of $D^{\times}$ where $D$ is a division
ring (1955 [2]). Amitsur used class field theory to complete the
classification.

######  Lemma 88.

Let $G$ be a group of order $p^{2}$ where $p$ is a prime. Suppose $F$ is a
field of characteristic not equal to $p$. If $G$ is freely representable over
$F$ then $G$ is cyclic.

###### Proof.

Suppose $G$ is freely representable but not cyclic. This means that every
nonidentity element of $G$ is in a unique cyclic group of order $p$. Observe
that there are $k=(p^{2}-1)/(p-1)=p+1$ such cyclic groups. Let $\mathcal{C}$
be the collection of cyclic subgroups of $G$ of order $p$. Then

$\sum_{C\in\mathcal{C}}{\mathbf{N}}C=(k-1)\mathbf{1}+{\mathbf{N}}G=p\mathbf{1}+{\mathbf{N}}G.$

Since $p$ is nonzero in $F$, this contradicts Theorem 11. ∎

Suppose $D$ is a division ring and that $G$ is a finite subgroup of
$D^{\times}$. If $F$ is the prime subfield of $D$, then let $F(G)$ be the
$F$-span of $G$ in $D$. (Warning: $F(G)$ is analogous to the group ring
$F[G]$, but they are not the same since $G$ might not be linearly
independent.)

######  Lemma 89.

Let $D,G,F,F(G)$ be as above, and suppose $F$ is ${\mathbb{F}}_{p}$ for some
prime $p$. Then $F(G)$ is a finite division ring of order a power of $p$.

###### Proof.

Let $k$ be the size of a basis of $F(G)$ for scalar field $F$, and observe
that $F(G)$ has finite size $p^{k}$. Observe that $F(G)$ is closed under
multiplication, so we conclude that $F(G)$ is a subring of $D$. Next suppose
$a\in F(G)$ is nonzero. Then the map $x\mapsto ax$ is an injective map
$F(G)\to F(G)$ since $F(G)$ is contained in a division ring. Since $F(G)$ is
finite, this map is surjective, and $ab=1$ for some $b\in F(G)$. This implies
that $F(G)$ is a division ring. ∎

######  Lemma 90.

Let $D$ be a division ring whose prime field $F$ has prime characteristic $p$.
Then $D^{\times}$ has no elements of order $p$.

###### Proof.

Suppose $g\in D^{\times}$ has order $p$, and let $G$ be the group generated by
$g$. Observe that $G$ is a subgroup of $F(G)^{\times}$, and $F(G)^{\times}$
has order $p^{k}-1$ for some $k\geq 1$. So the order of $g\in F(G)^{\times}$
fails to divide the order of $F(G)^{\times}$, a contradiction. ∎

######  Corollary 91.

Let $G$ be a subgroup of $D^{\times}$ where $D$ is a division ring. Then every
subgroup of $G$ of order $p^{2}$, where $p$ is a prime, is cyclic.

###### Proof.

Let $H$ be a subgroup of $G$ of order $p^{2}$ where $p$ is a prime. By the
above lemma, we can assume that $p$ is not the characteristic of the prime
field $F$ of $D$. By Lemma 87, $H$ is freely representable over $F$. So $H$ is
cyclic by Lemma 88. ∎

######  Corollary 92.

Suppose $G$ is a finite subgroup of $F^{\times}$ where $F$ is a field. Then
$G$ is cyclic.

###### Proof.

By the previous corollary, every subgroup of $G$ of order $p^{2}$ is cyclic,
for any prime $p$. Since $G$ is Abelian, $G$ is cyclic by the finite structure
theorem of Abelian groups (or you can use the more elementary argument given
in the remark after Corollary 19). ∎

######  Proposition 93.

Let $D$ be a division algebra and let $G$ be a finite subgroup of
$D^{\times}$. Then $G$ is a Sylow-Cycloidal group

###### Proof.

If $p$ is odd, then any $q$-Sylow subgroup of $G$ is cyclic by Theorem 25. If
$p=2$ then any $q$-Sylow subgroup is either cyclic or a generalized quaternion
group by Corollary 37.

Now suppose $F$ has prime characteristic $p$. In this case $G$ is a subgroup
of $F(G)^{\times}$, and $F(G)^{\times}$ has order $p^{k}-1$ for some $k\geq
1$. So $p$ cannot divide the order of $G$, so there are no $p$-Sylow subgroups
of $G$ that we need to worry about.

We conclude that all Sylow-subgroups of $G$ have the desired form, and that
$G$ is a Sylow-cycloidal group. ∎

######  Lemma 94.

Let $D$ be a division algebra of prime characteristic $p$, and let $G$ be a
finite subgroup of $D^{\times}$. Then $G$ is a Sylow-cyclic group

###### Proof.

Let $q$ be any odd prime dividing the order of $G$. By Corollary 91 and
Theorem 25, the $q$-Sylow subgroups of $G$ are cyclic.

Suppose the 2-Sylow subgroup of $G$ are not also cyclic. By Corollary 91 and
Proposition 37 the $2$-Sylow subgroups of $G$ must be generalized quaternion
groups. By Proposition 41, $G$ must then contain a subgroup $Q$ that we can
identify with the quaternion group of size 8. By Lemma 89 we have the division
algebra ${\mathbb{F}}_{p}(Q)$. The idea of the remainder of the proof is to
argue that there cannot be a “quaternion ring” over ${\mathbb{F}}_{p}$.

To proceed we solve $1+x^{2}+y^{2}=0$ over ${\mathbb{F}}_{p}$. If $p=2$ then
$x=1,y=0$ is a solution. Otherwise, observe that there are $(p+1)/2$ squares
in ${\mathbb{F}}_{p}$. So as $x$ varies in ${\mathbb{F}}_{p}$, the expression
$-1-x^{2}$ takes on $(p+1)/2$ distinct values. At least one of these values
must be a square since there are only $(p-1)/2$ nonsquares in
${\mathbb{F}}_{p}$. So we choose $x$ so that $-1-x^{2}$ is a square, and we
choose $y$ so that $y^{2}$ is $-1-x^{2}$. We can exchange $x$ and $y$ if
necessary, and assume $y\neq 0$. Then in ${\mathbb{F}}_{p}(Q)$ we have

$(1+x\mathbf{i}+y\mathbf{j})(1-x\mathbf{i}-y\mathbf{j})=1+x^{2}+y^{2}=0.$

Since ${\mathbb{F}}_{p}(Q)$ has no zero divisors, we have
$1+x\mathbf{i}+y\mathbf{j}=0$ or $1-x\mathbf{i}-y\mathbf{j}=0$. Since $y\neq
0$ this means that $\mathbf{j}\in{\mathbb{F}}_{p}(\left<\mathbf{i}\right>)$.
But ${\mathbb{F}}_{p}(\left<\mathbf{i}\right>)$ is a field, and so
$\mathbf{i}$ and $\mathbf{j}$ commute, a contradiction. ∎

The following is a result of Herstein. He proved it as a corollary of
Wedderburn’s theorem, but we will prove Wedderburn’s theorem as a corollary of
this result.

######  Theorem 95 (Herstein).

Let $D$ be a division ring of prime characteristic $p$, and let $G$ be a
finite subgroup of $D^{\times}$. Then $G$ is cyclic.

###### Proof.

First we consider the case where $D$ is finite and $G=D^{\times}$. By the
above lemma $G$ is a Sylow-cyclic group. Let $C$ be the maximum cyclic
characteristic (MCC) subgroup of $G$. Such a subgroup $C$ exists by Corollary
63 and, by Corollary 70, $C$ is maximal among cyclic subgroups of $G$. Observe
that ${\mathbb{F}}_{p}(C)$ is a field, so ${\mathbb{F}}_{p}(C)^{\times}$ is
cyclic. By the maximality of $C$, this means that
$C={\mathbb{F}}_{p}(C)^{\times}$.

Let $q$ be the number of elements of ${\mathbb{F}}_{p}(C)$, and let $q^{k}$ be
the number of elements of $D$. (Here we use the fact that $D$ is a vector
space over any subfield). If $k>1$ then

$|G/C|=\frac{q^{k}-1}{q-1}=q^{k-1}+\ldots+q+1\geq q+1>q-1=|C|$

which contradicts Corollary 72. Thus $k=1$ and so $G=C$, and $G$ is cyclic.

In general, we consider ${\mathbb{F}}_{p}(G)$. Since
${\mathbb{F}}_{p}(G)^{\times}$ is cyclic, as we have just shown, and since $G$
is a subgroup of ${\mathbb{F}}_{p}(G)^{\times}$, we conclude that $G$ is
cyclic as well. ∎

######  Corollary 96 (Wedderburn).

Every finite division ring $D$ is a field.

###### Proof.

By the above theorem $D^{\times}$ is cyclic. So $D$ must be a commutative
ring. ∎

##  Chapter 8 Sylow-Cycloidal Groups: The Solvable Case

Suppose $G$ is a solvable Sylow-cycloidal group and $O(G)$ is the maximal
normal subgroup of $G$ of odd order. Then our first important result will be
to describe the possible quotients $G/O(G)$. We will show that $G/O(G)$ is
isomorphic to either a cyclic $2$-group, a generalized quaternion group, the
binary tetrahedral group $2T$ or the binary octahedral group $2O$. In
particular, $G/O(G)$ is isomorphic to a solvable subgroup of
${\mathbb{H}}^{\times}$. This result divides Sylow-Cycloidal groups into four
mutually exclusive types. We then focus on each type individually.111I learned
the technique of classifying freely representable groups $G$ by their
quotients $G/O(G)$ from a recent paper by Daniel Allcock [1]. My approach
generalizes the scope of Allcock a bit. Although my proof is different and
more elementary that that in [17], I also lean on Wolf [17], and thus
indirectly on Zassenhaus (1936), for guidance. Zassenhaus adopts a more
general scope than mine by only restricting the $2$-Sylow subgroup (see Lemma
6.1.9 of Wolf [17] attributed to Zassenhaus).

###  1 The Quotient $G/O(G)$

We start more generally than with Sylow-cycloidal groups. We say a finite
group $G$ _satisfies the $(2,3)$ condition_ if every $2$-Sylow subgroup of $G$
is cyclic or is the quotient of a generalized quaternion group222This means
that the $2$-Sylow subgroups of $G$ are cyclic, dihedral, or generalized
quaternion, but we will not need this fact. and every $3$-Sylow subgroup is
cyclic.

######  Lemma 97.

Let $G$ be a group that satisfies the $(2,3)$ condition. Then every subgroup
and quotient of $G$ satisifies the $(2,3)$ condition.

###### Proof.

Observe that every $p$-Sylow subgroup of a quotient $G/N$ is a quotient of a
$p$-Sylow subgroup of $G$. Also every $p$-Sylow subgroup of a subgroup $H$ of
$G$ is a $p$-group and so is a subgroup of a $p$-Sylow subgroup of $G$.

The class of cyclic $p$-groups is closed under the processes of quotient and
subgroup. Since the class of cyclic and generalized quaternion $2$-groups is
closed under subgroup, the class of quotients of such groups is closed under
quotient and subgroups. ∎

######  Definition 7.

Let $G$ be a finite solvable group and let
$G,G^{\prime},G^{\prime\prime},\ldots,G^{(k)}=\\{1\\}$ be the derived series
of commutator subgroups. Then the _characteristic Abelian subgroup of $G$_,
which we denote as $\mathcal{A}(G)$, is defined to be the first Abelian term
of the series. Observe that $\mathcal{A}(G)$ is Abelian and characteristic,
and if $G$ is nontrivial then $\mathcal{A}(G)$ is also nontrivial.

######  Lemma 98.

Let $G$ be a solvable group satisfying the $(2,3)$ condition. Then either $G$
has order of the form $2^{m}3^{n}$ or there is a prime $p\geq 5$ such that
there is a $p$-subgroup $K$ of $G$ that is a characteristic subgroup of $G$.

###### Proof.

Let $q$ be a prime dividing $\mathcal{A}(G)$ (if no such $q$ exists then $G$
has order $2^{0}3^{0}$ and we are done). If $q\geq 5$ then we can choose $p=q$
and choose $K$ to be the $p$-Sylow subgroup of $\mathcal{A}(G)$, and we are
done. Otherwise we define an elementary characteristic subgroup $N$ as the
solutions of $x^{q}=1$ in $\mathcal{A}(G)$.

Since $G$ satisfies the $(2,3)$ condition, the same is true of $N$ and $G/N$.
If $q=3$ this means that $N$ is cyclic of order $3$. If $q=2$ then $N$ is
generated by one or two elements, so is either cyclic of order $2$ or is the
Klein four group. So any automorphism of $N$ has order $1,2,$ or $3$.

If $G/N$ has order of the form $2^{m}3^{n}$ we are done, so we will now assume
that $G/N$ is not of that form. By induction we can assume that $G/N$ has a
$p$-subgroup $L/N$ where $p\geq 5$ is prime and where $L$ is a subgroup of $G$
containing $N$ such that $L/N$ is a characteristic subgroup of $G/N$. Since
$N$ is a characteristic subgroup, this implies that $L$ must also be a
characteristic subgroup of $G$. Let $K$ be a $p$-Sylow subgroup of $L$ and
observe that $K$ is a complement of $N$ in $L$, so $L\cong N\rtimes K$. Note
that $K$ acts trivially on $N$ since all automorphism of $N$ have order prime
to $p$. This means that $L$ is isomorphic to $N\times K$, and so $K$ is the
unique $p$-Sylow subgroup of $L$. Since $L$ is a characteristic subgroup of
$G$, its unique $p$-Sylow subgroup $K$ is also characteristic subgroup of $G$.
∎

######  Corollary 99.

Let $G$ be a solvable group satisfying the $(2,3)$ condition. Let $O_{6}(G)$
be the maximal normal subgroup of $G$ of order relatively prime to $6$. Then
$G/O_{6}(G)$ has order of the form $2^{m}3^{n}$.

######  Corollary 100.

Let $G$ be a solvable group satisfying the $(2,3)$ condition. Let $O(G)$ be
the maximal normal subgroup of $G$ of odd order. Then $G/O(G)$ has order of
the form $2^{m}3^{n}$, and every nontrivial normal subgroup of $G/O(G)$ has
even order.

Motivated by the above corollary, we focus on Sylow-Cycloidal groups whose
order is of the form $2^{m}3^{n}$. Of course we know what such group are when
$m=0$ or when $n=0$, so we focus on the case where $m$ and $n$ are positive.

######  Lemma 101.

Let $G$ be a solvable Sylow-cycloidal group of order $2^{m}3^{n}$ with $m$ and
$n$ positive. Assume also that every normal subgroup of $G$ is of even order.
Then the following hold

  * •

$G$ contains a normal subgroup $N$ isomorphic to the quaternion group $Q_{8}$.

  * •

Every element of $G$ of order $3$ acts nontrivially on $N$.

  * •

$G$ contains four subgroups of order $3$, and the action of $G$ on these
subgroups gives a homomorphism $G\to S_{4}$ with kernel $Z(G)$.

  * •

The center $Z(G)$ has two elements.

  * •

$G/Z(G)$ is isomorphic to either $A_{4}$ or $S_{4}$, and so $G$ has order 24
or 48.

###### Proof.

Let $N$ be a maximal normal $2$-subgroup (i.e., the intersection of the
$2$-Sylow subgroups of $G$). Observe that characteristic Abelian subgroup
$\mathcal{A}(G/N)$ of $G/N$ must be a nontrivial $3$-group since if it had
order divisible by $2$ then we could violate the maximality of $N$ via the
$2$-Sylow subgroup of $\mathcal{A}(G/N)$. Let $H/N$ be the unique cyclic group
of order $3$ in $\mathcal{A}(G/N)$. Here $H$ is a subgroup of $G$ containing
$N$, and since $H/N$ is characteristic in $\mathcal{A}(G/N)$, we see that
$H/N$ is characteristic in $G/N$. Since $N$ is characteristic in $G$ we then
see that $H$ is characteristic in $G$. Let $C_{3}$ be a cyclic subgroup of $H$
of order $3$. Then we have that $H=NC_{3}$ is a semidirect product $N\rtimes
C_{3}$. If $C_{3}$ acts trivially on $N$ then $H$ would be isomorphic to
$N\times C_{3}$ and would contain a characteristic subgroup of order 3. This
violates our assumption on normal subgroups of $G$. So $C_{3}$ acts
nontrivially on $N$.

Up to isomorphism the only $2$-Cycloidal group with an automorphism of order 3
is the quaternion group $Q_{8}$, so $N$ is isomorphic to $Q_{8}$. Thus $H$ has
$24$ elements. Denote by $-1$ the unique element of $N$ of order $2$, which
must be the unique element of $H$ of order 2. Note that $-1$ is fixed by the
action of $C_{3}$, and so must be in the center of $H=NC_{3}$. In fact, the
action of $C_{3}$ on the quaternions only fixes $\pm 1$ so the center $Z(H)$
is $\\{\pm 1\\}$.

Let $L$ be the normalizer of $C_{3}$ in $H$. Observe that $L\cap N$ and
$C_{3}$ are normal in $L$, and so $L$ is isomorphic $(L\cap N)\times C_{3}$.
This means $C_{3}$ is characteristic in $L$. So $L$ cannot be normal in $H$,
otherwise $C_{3}$ would be normal in $H$, and hence in $G$, a contradiction.
So $L$ must have index at least $4$ in $H$. However, $L$ contains $C_{3}$ and
$Z(H)$. Thus $L$ has order $6$. Since $L$ is isomorphic to $(L\cap N)\times
C_{3}=Z(H)\times C_{3}$, it is cyclic of order $6$. By the orbit-stabilizer
theorem (and the fact that $p$-Sylow groups are conjugate) we get that $H$ has
exactly four $3$-Sylow subgroups. This action gives a homomorphism $H\to
S_{4}$. The kernel of this action is the intersection of the normalizers. Our
description of $L$ applies to all the normalizers, and the intersection is
seen to be $Z(H)$. So $Z(H)$ is the kernel of the action $H\to S_{4}$. Observe
that the image of $L$ under $H\to S_{4}$ is isomorphic to $L/Z(H)$, so is
cyclic of order $3$, and it fixes the element corresponding to $C_{3}$. This
observation applies not just to $C_{3}$ but to all 3-Sylow subgroups of $H$.
So the image of $H\to S_{4}$ contains all subgroup of order $3$. This means
that the image contains all of $A_{4}$, and must in fact be $A_{4}$ since
$H/\\{\pm 1\\}$ is a group of size $12$.

If $G=H$, then we are done. So for the remainder of the proof we assume $H$ is
not all of $G$. We claim that $G$ has has order $2^{m}3$ where $m>3$.
Otherwise every element $h$ of order $3$ in $H$ is of the form $g^{3}$ for
some element of $g$ of order $9$. Since such an $g$ acts on $N$ as an
automorphism of order 1 or 3, this forces $h$ to act trivially on $N$, a
contradiction to a previous conclusion.

The $2$-Sylow subgroups of $G$ cannot be normal in $G$ by maximality of $N$.
So there are three $2$-Sylow subgroups $S_{1},S_{2},S_{3}$ since the number of
such groups must divide $|G|/2^{m}$. Since $G$ acts transitively on
$\\{S_{1},S_{2},S_{3}\\}$ by conjugation, we get a homomorphism $G\to S_{3}$
whose image is $S_{3}$ or $A_{3}$. The kernel is a normal $2$-subgroup of $G$
of index 3 or 6 in $G$. By maximality of $N$, this kernel is contained in $N$.
So the index of $N$ in $G$ is a divisor of $6$, and since it has size at least
$6$ it is equal to $6$. Hence (1) $G$ has size $48$, (2) each $S_{i}$ is a
generalized quaternion group of size 16, (3) $H$ is the kernel of $G\to
S_{3}$, (4) the homomorphism $G\to S_{3}$ is surjective with each $S_{i}$
mapping to a different subgroup of order $2$ in $S_{3}$, and (5) $H$ has index
2 in $G$ and its image under $G\to S_{3}$ is $A_{3}$.

Observe that every $3$-Sylow subgroup of $G$ is contained in $H$ since
$[G:H]=2$. Thus the set of $3$-Sylow subgroups of $G$ has size 4, giving a
homomorphism $G\to S_{4}$ extending the earlier surjection $H\to A_{4}$. This
means that the kernel $K$ of $G\to S_{4}$ has order $2$ or $4$. The image of
$K$ under $G\to S_{3}$ is a normal $2$-subgroup of $S_{3}$, so $K$ has trivial
image in $S_{3}$. This means that $K\subseteq H$. But we already know that the
kernel of $H\to S_{4}$ is $Z(H)$. Hence $K=Z(H)$. This implies that $G/Z(H)$
is isomorphic to $S_{4}$.

We conclude by showing that $Z(H)=Z(G)$. First observe that under the map
$G\to S_{4}$ the center $Z(G)$ maps to the center of $S_{4}$ which is trivial.
So $Z(G)$ is contained in the kernel $Z(H)$. Conversely, since $S_{1}\cap
S_{2}\cap S_{3}$ is the maximal normal 2-subgroup of $G$, this intersection is
$H$. Thus $Z(H)\subseteq S_{i}$, and so $Z(H)$ is the unique subgroup of
$S_{i}$ of size 2. So $Z(H)$ is contained in $Z(S_{i})$ for each $i$. Since
$S_{1},S_{2},S_{3}$ generates $G$ we have that $Z(H)\subseteq Z(G)$. ∎

######  Remark.

By a theorem of Burnside, all groups of order $p^{m}q^{n}$ are solvable for
distinct primes $p$ and $q$. So accepting this result allows us to drop the
solvability assumption for groups of order $2^{m}3^{n}$. I will keep this
assumption just to make the proofs a bit more accessible.

There are two cases in the above lemma. We now link these cases to specific
subgroups of ${\mathbb{H}}^{\times}$.

######  Proposition 102.

If $G$ is a solvable Sylow-cycloidal group of order 24 with no normal subgroup
of order 3, then $G$ is isomorphic to the binary tetrahedral group $2T$.

###### Proof.

By the above lemma, we can identify $Q_{8}$ with a normal subgroup of $G$.
Also, $G$ contains four subgroups of order $3$, each which acts nontrivially
on $Q_{8}$, and the center of $G$ has two elements. Note that we have an
embedding of $G/Z(G)$ into ${\mathrm{Aut}}(Q_{8})$. This implies that if
$\alpha$ and $\beta$ are distinct elements of order 3, their actions on
$Q_{8}$ are distinct (otherwise $\alpha=\epsilon\beta$ where $\epsilon\in
Z(G)$, and $\epsilon$ must be $1$ since $\alpha$ has order $3$).

Note that any automorphism $\phi$ of order $3$ of $Q_{8}$ must permute its
three subgroups of order 4, and cannot fix any such subgroup (otherwise it
would have to fix all three, and then $\phi^{2}$ would fix all of $Q_{8}$). So
there are only 4 possible values of $\phi(\mathbf{i})$, and they are contained
in $\\{\pm\mathbf{j},\pm\mathbf{k}\\}$. Once $\phi(\mathbf{i})$ is known,
there are only two possibilities for $\phi(\mathbf{j})$. Since $\mathbf{i}$
and $\mathbf{j}$ generate $Q_{8}$, this gives $8$ possibilities. All of these
possibilities must occur as the automorphism associated with elements of order
3, since there are 8 such elements.

Thus we can find an element $g\in G$ of order 3 such that it acts on $Q_{8}$
by sending $\mathbf{i}$ to $\mathbf{j}$ and sending $\mathbf{j}$ to
$\mathbf{k}$. Let $C_{3}$ be the subgroup generated by $g$. Then

$G\cong Q_{8}\rtimes C_{3}$

where $C_{3}$ acts on $G$ as specified. Thus $G$ is unique up to isomorphism.

Since $2T$ is freely representable, it is a Sylow-cycloidal group. Also
$2T/\\{\pm 1\\}$ is isomorphic to $A_{4}$. Since $A_{4}$ is solvable and has
no normal subgroups of order 3, the same is true of $2T$. So $2T$ satisfies
our assumptions for $G$, and so must be isomorphic to any such $G$. ∎

Here is a variant of the above:

######  Corollary 103.

Let $G$ be a finite group of order 24 that has a unique element of order $2$
but does not have a unique subgroup of order $3$. Then $G$ is isomorphic to
the binary tetrahedral group $2T$.

###### Proof.

Let $P_{2}$ be a $2$-Sylow subgroup of $G$. Then $P_{2}$ has a unique element
of order $2$, so must be isomorphic to $Q_{8}$ or a cyclic group of order $8$.
In particular, $G$ is a Sylow-cycloidal group. Note that the subgroups of
order $3$ of $G$ are conjugate since they are Sylow subgroups, so they cannot
be normal since there is more than one. The result now follows from
Proposition 98 if we grant that $G$ is solvable.

It is well-known that all subgroups of order 24 are solvable, but to see this
directly for $G$ here let $C_{2}$ be the unique subgroup of order $2$ and let
$C_{3}$ be any subgroup of order $3$. Then the subgroup
$C_{2}C_{3}=C_{2}\rtimes C_{3}$ must be cyclic since $C_{2}$ has no
automorphism. So the normalizer of $C_{3}$ has at least 6 elements. This means
that there are at most $4$ subgroups of $G$ of order $3$. Since the number of
such Sylow subgroups is congruent to 1 modulo $3$, we see there are exactly
$4$ subgroups of order $3$ and $C_{2}C_{3}$ is the normalizer of $C_{3}$. The
action of $G$ on the set of subgroups of order $4$ gives a homomorphism $G\to
S_{4}$. The kernel of the action is the intersection of the normalizers for
subgroups of order 3. Our description of the normalizer shows this
intersection is $C_{2}$. Since $S_{4}$ is solvable, this implies that $G$ is
solvable. ∎

######  Proposition 104.

Let $G$ be a solvable Sylow-cycloidal group of order 48 such that $G/Z(G)$ is
isomorphic to $S_{4}$. Then $G$ is isomorphic to the binary octahedral group
$2O$. Furthermore $G$ has $8$ elements of order 3, and $4$ subgroups of order
3, and these elements generate the unique subgroup of $G$ of index $2$, and
this index 2 subgroup is a binary tetrahedral group.

###### Proof.

We start by checking that $G=2O$ satisfies the hypothesis. Recall that $2O$
has $48$ elements. We note that $2O$ has center containing the elements $\pm
1$, and since $2O/\\{\pm 1\\}\cong O\cong S_{4}$ the center is exactly $\\{\pm
1\\}$ (since $S_{4}$ has trivial center). So $2O/Z(2O)$ is isomorphic to
$S_{4}$.

Now let $G$ be any subgroup of size 48 such that $G/Z(G)\cong S_{4}$. Observe
that $Z(G)$ has order $2$ since $G$ has order 48 and $S_{4}$ has order $24$.
So the generator of $Z(G)$ is the unique element of $G$ of order 2 (Lemma 49).
For now, we fix a particular isomorphism $G/Z(G)\to S_{4}.$ Since $Z(G)$ has
order $2$, the order of the image of $g\in G$ in $S_{4}$ is either the same as
the order of $g$ or has half the order of $g$. For example if $g$ maps to an
element of order 2 then $g$ must have order 2 or 4. But in this case $g$ is
not the unique element of order 2 since that maps to the identity element, so
$g$ has order $4$ and $g^{2}$ is the unique element of order $2$.

In contrast if $g$ has order 3, then $g$ must map to an element of order $3$
since $3$ is odd. In particular if $g$ has order 3 then $g$ maps to a 3 cycle
in $S_{4}$. So if $\mathcal{S}_{3}$ is the set of elements of order $3$ in $G$
then we have a map $\mathcal{S}_{3}\to\mathcal{T}_{3}$ where $\mathcal{T}_{3}$
is the set of three cycles. Now let $t\in\mathcal{T}_{3}$ be given. The
subgroup $\left<t\right>$ of $S_{4}$ corresponds to a subgroup $H_{t}$ of $G$
of order $6$ containing $Z(G)$ as a normal subgroup. In fact,
$H_{t}=Z(G)C=Z(G)\rtimes C$ where $C$ is a 3-Sylow subgroup of $H_{t}$. Since
$Z(G)$ has no nontrivial automorphisms, $H_{t}$ is just $Z(G)\times C$, so $C$
is the unique subgroup of $G$ order 3 whose image in $S_{3}$ is
$\left<t\right>$. Thus there is exactly one element of $G$ that maps to $t$.
This means that $\mathcal{S}_{3}\to\mathcal{T}_{3}$ is a bijection. In
particular, there are $8$ elements of order 3 in $G$, and $4$ subgroups of
order 3 in $G$.

Let $H$ be the subgroup of $G$ generated by the set $\mathcal{S}_{3}$ of
elements of order 3. The image in $S_{4}$ is $A_{4}$ (since $\mathcal{T}_{3}$
generates $A_{4}$). This implies that $H$ is also of even order and so must
contain the unique subgroup $Z(G)$ of $G$ of order $2$. Thus $H/Z(G)\cong
A_{4}$, and so $H$ has order $24$. Also if $H$ has a normal subgroup of size
3, then its image in $A_{4}$ would also have a normal subgroup of size 3. But
this is not the case. So we have that $H$ is isomorphic to the binary
tetrahedral group $2T$ (see previous proposition). We also note that $H$ is
the only subgroup of $G$ of order 24. So see this note that any other such
group $H^{\prime}$ would have to contain the unique element of $G$ order $2$
and so would contain $Z(G)$. Thus its image in $S_{4}$ would have size 12. But
$A_{4}$ is the only subgroup of $S_{4}$ of index two333A subgroup $N$ of
$S_{4}$ of index 2 is normal, and contains all three cycles since $G/N$ has
order 2. Since three cycles generate $A_{4}$ any such $N$ would have to be
$A_{4}$. and so $H^{\prime}$ and $H$ would have the same image in $S_{4}$ and
so would be equal.

This gives us enough information to describe $G$ in terms of relations. The
final specification of $G$ will not depend on a particular isomorphism
$G/Z(G)\cong S_{4}$ but only on the fact that such an isomorphism exists. Note
that the subgroup $H$ (the unique subgroup of $G$ of index 2) and the subset
$\mathcal{S}_{3}$ do not depend on the map. Choose an element $g_{1}\in G$ of
order $3$. It can be any such element. Next choose $g_{2}$ to be any element
of order $3$ such that the product $g_{1}g_{2}$ has order 4. To show this can
be done it is useful to choose a map $G/Z(G)\cong S_{4}$ (by permuting the
number if necessary from a given map) so that $g_{1}$ corresponds to
$(1\,2\,3)$. Then the element $g_{2}$ corresponding to $(1\,2\,4)$ will work
since $g_{1}g_{2}$ maps to $(1\,3)(2\,4)$ of order $2$ in $S_{4}$ and so
$g_{1}g_{2}$ has order $4$ in $G$. In fact, once we have chosen a suitable
$g_{1}$ and $g_{2}$, we can permute the numbering of the four elements
permuted by $S_{4}$ so that $g_{1}$ corresponds to $(1\,2\,3)$ and $g_{2}$
corresponds to $(1\,2\,4)$. Note that $(34)(1\,2\,3)(34)=(1\,2\,4)$ so if we
choose $\tau\in G-H$ mapping to $(34)$ then $\tau g_{1}\tau^{-1}$ is an
element of order 3 corresponding to $(1\,2\,4)$. Since
$\mathcal{S}_{3}\to\mathcal{T}_{3}$ is a bijection, this means that

$\tau g_{1}\tau^{-1}=g_{2}.$

We also have

$\tau g_{1}^{-1}\tau^{-1}=g_{2}^{-1},\quad\tau g_{2}\tau^{-1}=g_{1},\quad\tau
g_{1}^{-1}\tau^{-1}=g^{-1}_{1}.$

Here we made use of $\tau^{-1}g\tau=\tau g\tau^{-1}$ for all $g\in G$ since
$\tau^{2}$ has order 2 in $Z(G)$. By conjugating other 3-cycles in $S_{4}$ by
$(3\,4)$ we can see that

$\tau g\tau^{-1}=g^{-1}$

for the other four elements $g\in\mathcal{S}_{3}$. This gives us eight
relations, one for each element of $\mathcal{S}_{3}$. We verified they held by
using a particular map $G/Z(G)\cong S_{4}$, but the actual relations
themselves do not depend on the map. In addition we have a ninth relation
$\tau^{2}=-1$ where $-1$ is the unique element of order $2$ in $H$.

Now consider the free product $H*C_{4}$ where $C_{4}$ is an abstract cyclic
group of order $4$ with generator called $\tau$. Let $K$ be the normal
subgroup generated by the 9 relations discussed above. Note that every element
of $H*C_{4}/K$ can be written as $a$ or $a\overline{\tau}$ where $a$ is in the
image of $H$ and $\overline{\tau}$ is the image of $\tau$. This follows from
the fact that $\mathcal{S}_{3}$ generates $H$. Thus the group $H*C_{4}/K$ has
at most $48$ elements. However, $G$ satisfies these relations so there is a
homomorphism $H*C_{4}/K\to G$, and this is a surjection since $G$ is generated
by $\tau$ and $H$. Thus $G$ is isomorphic to $H*C_{4}/K$.

So if $G_{1}$ and $G_{2}$ are solvable Sylow-cycloidal groups of order 48 that
contain a common subgroup $H$ of size 24, and if $G_{1}/Z(G_{1})$ and
$G_{2}/Z(G_{2})$ are both isomorphic to $S_{4}$, then $G_{1}$ must be
isomorphic to $G_{2}$ since both are isomorphic to $H*C_{4}/K$ (and $K$ does
not depend on $G_{i}$ but only on $H$). More generally, if $G_{1}$ and $G_{2}$
are are solvable Sylow-cycloidal groups of order 48 such that $G_{1}/Z(G_{1})$
and $G_{2}/Z(G_{2})$ are isomorphic to $S_{4}$, then as observed above each
$G_{i}$ has a subgroup isomorphic to $2T$. By identifying these subgroups we
reduce to the situation where $G_{1}$ and $G_{2}$ share a subgroup of order
$24$. We conclude that $G_{1}$ and $G_{2}$ are isomorphic under these
conditions. ∎

Here are some useful observations linking the $2$-Sylow subgroup of $G$ to the
quotient $G/O(G)$.

######  Proposition 105.

Let $G$ be finite group and let $O(G)$ be its maximal normal subgroup of odd
order. Then $G$ and $G/O(G)$ have isomorphic $2$-Sylow subgroups.

######  Corollary 106.

Let $G$ be a solvable Sylow-cycloidal group whose $2$-Sylow subgroup $S$ is
not quaternionic of order 8 or 16. Then $G/O(G)$ is isomorphic to $S$.

######  Corollary 107.

Let $G$ be a Sylow-cycloidal group. Then $G$ is a Sylow-cyclic group if and
only if $G/O(G)$ is a cyclic $2$-group.

So we divide the solvable Sylow-cycloidal groups $G$ into four mutually
exclusive types:

  1. 1.

Sylow-cyclic groups. These are the Sylow-cycloidal groups where $G/O(G)$ is
cyclic.

  2. 2.

Quaternion type. These are defined to encompass the Sylow-cycloidal groups
where $G/O(G)$ is a generalized quaternion group.

  3. 3.

Binary tetrahedral type. These are defined to encompass the Sylow-cycloidal
groups where $G/O(G)\cong 2T$.

  4. 4.

Binary octahedral type. These are defined to encompass the Sylow-cycloidal
groups where $G/O(G)\cong 2O$.

In addition, there are non-solvable Sylow-cycloidal groups that we will
consider later. These are Sylow-cycloidal groups that contain a perfect Sylow-
cycloidal subgroup.

###  2 Type 1: Sylow-cyclic groups

As noted out above, a Sylow-cycloidal group $G$ is of this type if and only if
$G/O(G)$ is a cyclic $2$-group. Such groups were treated in Section 6. In
particular, this type of group $G$ is freely representable if and only if it
has a unique subgroup of order $p$ for each prime $p$ dividing the order of
$G$.

In order to compare with later results it is convenient to break out the odd
part from the even part:

######  Proposition 108.

Let $G$ be a Sylow-cyclic group of order $2^{k}n$ where $n$ is odd and where
$k\geq 1$. Then $G$ has a normal Sylow-cyclic subgroup $M$ of order $n$. For
such $M$, the group $G$ is freely representable if and only if (1) $M$ is
freely representable and (2) $G$ has a unique element of order $2$.

###### Proof.

Let $M=O(G)$, so $G/M$ is a cyclic $2$-group (Corollary 107), and $O(G)$ has
odd order by definition, so $M$ has order $n$.

If $G$ is freely representable then (1) and (2) hold by earlier results. So
assume (1) and (2). To show $G$ is freely representable, it is enough to show
that $G$ has a unique subgroup of order $p$ for each $p$ dividing $2^{k}n$.
For $p=2$ we are covered by assumption (2). For an odd prime $p$ we note that
every subgroup of $G$ of order $p$ is a subgroup of $M$ since $G/M$ is a
$2$-group. By (1) there is a unique subgroup of $M$ (and hence of $G$) of
order $p$. ∎

###  3 Case 2: Quaternion Type

As noted in Proposition 105, these groups have $2$-Sylow subgroups that are
generalized quaternion groups, and if conversely if $G$ is a solvable Sylow-
cycloidal group with $2$-Sylow subgroups that are generalized quaternion
groups of order $32$ or more then $G$ must be of this type. (If $G$ is a
solvable Sylow-cycloidal group with $2$-Sylow subgroups isomorphic to the
quaternion group $Q_{8}$ of order $8$, it can either be of this type or of
binary tetrahedral type. If $G$ is a solvable Sylow-cycloidal group with
$2$-Sylow subgroups isomorphic to the generalized quaternion group $Q_{16}$ of
order $16$, it can either be of this type or of binary octahedral type.)

Observe that for Sylow-cycloidal groups $G$ of quaternion type, any $2$-Sylow
subgroup $Q$ of $G$ functions as a complement for the normal subgroup $O(G)$
of $G$. Thus

$G=O(G)Q=O(G)\rtimes Q.$

In particular, up to isomorphism $G$ is determined by $O(G)$ and the action of
$Q$ on $O(G)$.

######  Proposition 109.

Let $G$ be a Sylow-cycloidal group of quaternion type. Then $G$ has a unique
element of order 2. This element is in the center of $G$.

###### Proof.

Let $Q$ be a $2$-Sylow subgroup of $G$, which is a generalized quaternion
group (Proposition 105). We start by considering the action of $Q$ on $G$. By
Proposition 86, we have that $A={\mathrm{Aut}}(O(G))/O({\mathrm{Aut}}(O(G)))$
is an Abelian $2$-group. The action of $Q$ gives a map into $A$:

$Q\to{\mathrm{Aut}}(O(G))\to{\mathrm{Aut}}(O(G))/O({\mathrm{Aut}}(O(G)))=A.$

Since $A$ is Abelian, this map has a nontrivial kernel. Thus if $C_{2}$ is the
unique subgroup of $Q$ of order 2, then $C_{2}$ is in this Kernel. In
particular, the image of $C_{2}$ in ${\mathrm{Aut}}(O(G))$ must land in
$O({\mathrm{Aut}}(O(G)))$. But $O({\mathrm{Aut}}(O(G)))$ has odd order, so the
image of $C_{2}$ in ${\mathrm{Aut}}(O(G))$ is trivial. Thus $C_{2}$ acts
trivially on $O(G)$. Since $G=O(G)Q$ we have that $C_{2}$ is in the center of
$G$.

Observe that the subgroup $O(G)C_{2}$ of $G$ corresponds to the unique
subgroup of $G/O(G)$ of order two. If $g\in G$ has order 2, then its image in
$G/O(G)$ is the unique element of order 2, and so $g\in O(G)C_{2}$. Since
$C_{2}$ is in the center of $G$, we have $O(G)C_{2}=O(G)\times C_{2}$, and so
every element of $O(G)C_{2}$ of order 2 must be in $C_{2}$. We conclude that
every element of order $2$ is in $C_{2}$. In other words, there is a unique
element of order $2$ in $G$. ∎

Let $G$ be a Sylow-cycloidal group of quaternion type and let $C_{2}$ be its
unique subgroup of order 2. Let $M=O(G)C_{2}$. Observe that every element of
odd prime order must be in $O(G)$, so every element of $G$ of prime order is
in $M$. By Corollary 80, $G$ is freely representable if and only if $M$ is
freely representable. Since $M\cong O(G)\times C_{2}$ we have the $M$ is
freely representable if and only if $O(G)$ and $C_{2}$ are freely
representable (Corollary 21), but of course $C_{2}$ is freely representable.
Thus we get the following:

######  Theorem 110.

Let $G$ be a Sylow-cycloidal group of quaternion type. Then the following are
equivalent:

  1. 1.

$G$ is freely representable.

  2. 2.

$O(G)$ is freely representable.

  3. 3.

For each prime $p$ dividing the order of $G$, there is exactly one subgroup of
$G$ of order $p$.

###### Proof.

The equivalence $(1)\iff(2)$ was addressed in the discussion proceeding the
statement of the theorem. The implication $(3)\implies(1)$ follows from
Proposition 84.

So we just need to verify that $(2)$ implies $(3)$. We know that $(3)$ holds
for $p=2$ by the previous proposition (independent of whether $(2)$ is true or
not). So suppose that $(2)$ holds and that $p$ is an odd prime dividing the
order of $G$. Observe that every subgroup of order $p$ of $G$ must actually be
a subgroup of $O(G)$ since $G/O(G)$ has even order. Finally, $(2)$ implies
$O(G)$ has exactly one subgroup of order $p$ by Theorem 81. ∎

Now we consider other characterizations of this type of group.

######  Proposition 111.

Let $G$ be a finite group. Then $G$ is a Sylow-cycloidal group of quaternion
type if and only if it is a semidirect product $M\rtimes Q$ where $M$ is a
Sylow-cyclic group of odd order and $Q$ is a generalized quaternion group.

If $G$ is a semidirect product $M\rtimes Q$ where $M$ is a Sylow-cyclic group
of odd order and $Q$ is a generalized quaternion group, then $G$ is freely
representable if and only if $M$ is freely representable.

###### Proof.

We mentioned earlier that if $G$ is a Sylow-cycloidal group of quaternion type
then $G=O(G)\rtimes Q$ where $Q$ is any $2$-Sylow subgroup of $G$ and where
$Q$ is a generalized quaternion group. Conversely suppose $G$ is of the form
$M\rtimes Q$. Then $M$ is isomorphic to a normal subgroup of $G$ of odd order,
and the $2$-group $Q$ is isomorphic to the corresponding quotient of $G$. Thus
$O(G)$ must be isomorphic to $M$, and $G/O(G)$ is isomorphic to $Q$. So by the
above theorem, $G$ is freely representable if and only if $M$ is freely
representable. ∎

######  Proposition 112.

Let $G$ be a finite group. Then $G$ is a Sylow-cycloidal group of quaternion
type if and only _(1)_ the $2$-Sylow subgroups of $G$ are generalized
quaternion groups, and _(2)_ $G$ has a Sylow-cyclic subgroup $M$ of index $2$.
In this case $G$ is freely representable if and only if $M$ is freely
representable.

###### Proof.

Suppose is a Sylow-cycloidal group of quaternion type, so $G/O(G)$ is a
generalized quaternion group. This quotient contains a cyclic subgroup of
index 2 which we can write as $M/O(G)$ where $M$ is a subgroup of $G$
containing $O(G)$. Note that $M$ has index $2$ in $G$. Since $O(G)$ has odd
order, the $2$-Sylow subgroups of $M$ are isomorphic to the cyclic group
$M/O(G)$. Thus $M$ is a Sylow-cyclic subgroup of $G$. We also note that
$G/O(G)$ is isomorphic to the 2-Sylow subgroups of $G$, so these $2$-Sylow
subgroups of $G$ are generalized quaternion groups.

Conversely, suppose (1) and (2) hold. Since $M$ is a Sylow-cyclic group, the
quotient $M/O(M)$ is a cyclic $2$-group. Since $G/M$ has order 2, the subgroup
$O(G)$ must be a subgroup of $M$, so $O(G)\subseteq O(M)$ by the maximality of
$O(M)$. Since $M$ is normal in $G$ and since $O(M)$ is characteristic in $M$
it follows that $O(M)$ is normal in $G$. Thus $O(G)=O(M)$. Since
$M/O(M)=M/O(G)$ is a $2$-group and since $G/M$ is a $2$-group, it follows that
$G/O(G)$ is a $2$-group. Any Sylow $2$-group of $G$ is thus isomorphic to
$G/O(G)$, and so $G/O(G)$ is a generalized quaternion group. Also note that
since $G/M$ has order 2, all odd order Sylow-subgroups of $G$ are contained in
$M$, and so are cyclic. Thus $G$ is a Sylow-cycloidal group of quaternion
type.

If $G$ is freely representable, then so is the subgroup $M$. Conversely, if
$M$ is freely representable then so is $O(G)$ since, as noted above, $O(G)$ is
a subgroup of $M$. Thus $G$ is freely representable by Theorem 110. ∎

######  Proposition 113.

Let $G$ be a finite group and let $S$ be a $2$-Sylow subgroup of $G$. Then $G$
is a Sylow-cyclic group if and only if $S$ is cyclic and $G$ has a normal
Sylow-cyclic subgroup $M$ of index $|S|$. Similarly, $G$ is a Sylow-cycloidal
group of quaternion type if and only if $S$ is a generalized quaternion group
and $G$ has a normal Sylow-cyclic subgroup $M$ of index $|S|$.

###### Proof.

If $G$ is Sylow-cyclic group or a Sylow-cycloidal group of quaternion type,
then let $M=O(G)$ which is normal. Conversely, suppose there is a normal
Sylow-cyclic subgroup $M$ of index $|S|$. This group $M$ has odd order, and is
maximal with this property, so $M=O(G)$. It follows that $G/O(G)\cong S$. Note
also that every odd ordered Sylow subgroup of $G$ is actually in $M$ since
$G/M$ has even order. Thus every odd ordered Sylow subgroup of $G$ is cyclic.
∎

###  4 Case 3: Binary Tetrahedral Type

Let $G$ be a Sylow-cycloidal group of binary tetrahedral type and let $O(G)$
be its maximal normal subgroup of odd order. Then $G/O(G)$ is isomorphic to
the binary tetrahedral group $2T$, so the $2$-Sylow subgroups of $G$ are all
isomorphic to the quaternion group of order 8 (Proposition 105). What is
interesting about this case is that the $2$-Sylow subgroup of $G$ is unique,
and so is characteristic:

######  Proposition 114.

Let $G$ be a Sylow-cycloidal group of binary tetrahedral type. Then $G$ has a
unique $2$-Sylow subgroup of order $8$, and this $2$-Sylow subgroup is
isomorphic to the quaternion group with 8 elements. Furthermore, the $2$-Sylow
subgroup of $G$ centralizes $O(G)$.

###### Proof.

We start with the fact that $2T$ has a cyclic quotient $C_{3}$ of order 3. Let
$H$ be the kernel of the composition $G\to G/O(G)\cong 2T\to C_{3}$. Note that
every $2$-Sylow subgroup of $G$ must be in the kernel $H$. So we just need to
show that $H$ has a unique $2$-Sylow subgroup. Let $Q$ be a $2$-Sylow subgroup
of $G$, and observe that $Q$ is a complement for $O(G)$ in $H$. So

$H=O(G)Q=O(G)\rtimes Q.$

To prove the uniqueness result for $H$, and hence for $G$, it is enough to
show that $Q$ acts trivially on $O(G)$.

By Proposition 86, $A={\mathrm{Aut}}(O(G))/O({\mathrm{Aut}}(O(G)))$ is an
Abelian $2$-group. Let $G$ act on $O(G)$ by conjugation. Then we have the
composition

$G\to{\mathrm{Aut}}(O(G))\to{\mathrm{Aut}}(O(G))/O({\mathrm{Aut}}(O(G)))=A.$

Since the codomain is a $2$-group, we have $O(G)$ is in the Kernel. So we get
a map

$2T\cong G/O(G)\to A.$

The kernel must contain the commutator subgroup of $2T$. The commutator
subgroup of $2T$ contains the commutator subgroup of $Q_{8}$, so contains the
unique subgroup $C_{2}$ of $2T$ of order $2$. But $2T/C_{2}$ is isomorphic to
$T=A_{4}$, whose commutator subgroup is the normal subgroup of order $4$. Thus
the commutator subgroup $(2T)^{\prime}$ of $2T$ has index $3$ in $2T$. Thus we
get a homomorphism

$(2T)/(2T)^{\prime}\to A.$

Since $(2T)/(2T)^{\prime}$ has order 3, and $A$ is a $2$-group, we get that
the image of $G$ in $A$ is trivial. In other words, the image of $G$ in
${\mathrm{Aut}}(O(G))$ has odd order. In particular, the image of $Q$ in
${\mathrm{Aut}}(O(G))$ must be trivial. So $Q$ acts trivially on $O(G)$. ∎

######  Corollary 115.

Let $G$ be a Sylow-cycloidal group of binary tetrahedral type. Then $G$ has a
unique element of order $2$.

Let $M/O(G)$ be the subgroup of $G/O(G)$ of order $8$. Here $M$ is a subgroup
of $G$ containing $O(G)$, and since $M/O(G)$ is normal in $G/O(G)$ we have
that $M$ is a normal subgroup of $G$. Let $Q_{8}$ be the $2$-Sylow subgroup of
$G$. Observe that $Q_{8}$ is in $M$ since $G/M$ has order 3. In fact $Q_{8}$
is a complement for $O(G)$ in $M$, so

$M=O(G)Q_{8}=O(G)\times Q_{8}$

since $Q_{8}$ acts trivially on $O(G)$. We have that $Q_{8}$ is freely
representable, so $M$ is freely representable if and only if $O(G)$ is freely
representable (Corollary 21). The further analysis of freely representable
depends on whether or not $9$ divides the order of $G$.

######  Theorem 116.

Let $G$ be a Sylow-cycloidal group of binary tetrahedral type. If $9$ divides
the order of $G$ then the following are equivalent.

  1. 1.

$G$ is freely representable.

  2. 2.

$O(G)$ is freely representable.

  3. 3.

For each prime $p$ dividing the order of $G$, there is exactly one subgroup of
$G$ of order $p$.

###### Proof.

The property of being freely representable is inherited by subgroups so
$(1)\implies(2)$.

Now suppose that $(2)$ holds. Note that $(3)$ holds for $p=2$ (independent of
whether $(2)$ is true or not) by the above Corollary. Suppose that $C$ is a
subgroup of $G$ of order $p$ where $p$ is an odd prime. If $p\neq 3$ then $C$
is in $O(G)$ since $G/O(G)$ has order prime to $p$. If $p=3$, then $C$ is
contained in a $3$-Sylow subgroup $P$ of order at least $9$. The map $P$ to
$G/O(G)$ has image of size $3$, so it has a nontrivial kernel. Since $P$ is
cyclic, all nontrivial subgroups of $P$ contain $C$. So $C$ is in the kernel
of $P\to G/O(G)$. In other words, $C$ is in $O(G)$. Since all subgroups of odd
prime order of $G$ are in $O(G)$, we have uniqueness for each prime order by
Theorem 81. So $(3)$ holds.

Finally the implication $(3)\implies(1)$ follows from Proposition 84. ∎

######  Theorem 117.

Let $G$ be a Sylow-cycloidal group of binary tetrahedral type. If $9$ does not
divide the order of $G$ then $G$ has a subgroup $H$ isomorphic to $2T$ and

$G=O(G)H\cong O(G)\rtimes 2T.$

Furthermore, the following are equivalent.

  1. 1.

$G$ is freely representable.

  2. 2.

$O(G)$ is freely representable and $G\cong O(G)\times 2T$.

  3. 3.

For each prime $p\neq 3$ dividing the order of $G$ there is exactly one
subgroup of $G$ of order $p$, and there are $4$ subgroups of order $3$ in $G$.

###### Proof.

Let $C_{3}$ be a $3$-Sylow subgroup of $G$, and let $Q_{8}$ be the unique
$2$-Sylow subgroup of $G$. Then $Q_{8}$ is normal in $G$, so $H=Q_{8}C_{3}$ is
a subgroup of $G$ of order $24$. Note that $O(G)$ has order prime to $24$, so
$H$ maps isomorphically onto the quotient $G/O(G)$. Hence $H$ is isomorphic to
$2T$. Also observe that $H$ is a complement to $O(G)$ so

$G=O(G)H\cong O(G)\rtimes 2T.$

Next observe that $O(G)C_{3}=O(G)\rtimes C_{3}$ is a Sylow-cyclic subgroup of
$G$. So if $G$ is freely representable, then same is true of $L=O(G)C_{3}$. If
$L=O(G)C_{3}$ is freely representable then, by Theorem 81, $C_{3}$ is the only
subgroup of $L$ of order 3. Thus $C_{3}$ is normal in $L$, and so
$L=O(G)\times C_{3}$. In particular $C_{3}$ acts trivially on $O(G)$. Since
$H=Q_{8}C_{3}$ and since $Q_{8}$ acts trivially on $O(G)$ then $H$ acts
trivially on $O(G)$. So $(1)\implies(2)$ holds.

Note that $(2)\implies(1)$ by Corollary 21.

Next observe that if $G\cong O(G)\times 2T$ then every subgroup of prime order
$p\neq 2,3$ of $G$ must be in $O(G)$ and the subgroups of prime order $p=2$ or
$p=3$ correspond to the subgroup of $2T$ of prime order. If $O(G)$ is freely
representable then there is one subgroup of order $p$ for each $p\neq 2,3$
dividing the order of $G$ (and hence the order of $O(G)$) (Theorem 81).
Observe that $2T$ has one subgroup of order 2 and four subgroups of order 3.
So $(2)\implies(3)$.

Finally suppose $(3)$ holds. We can conclude that $O(G)$ is freely
representable by Proposition 84. Since $H$ itself has four subgroups of order
3, we have at least four subgroups of order $3$. Note the four subgroups of
$H$ of order $3$ map to distinct subgroups of $G/O(G)$. As before, let
$L=O(G)C_{3}$ where $C_{3}$ is a cyclic subgroup of $H$. Any subgroup of order
3 in $L$ maps to the same subgroup of $G/O(G)$ as $C_{3}$. Since there are
only 4 subgroups of order 3 in $G$ we conclude that $C_{3}$ is the unique
subgroup of order $3$ in $L$. So $C_{3}$ is normal in $L$, and $L=O(G)\times
C_{3}$. Thus $C_{3}$ acts trivially on $O(G)$. Since $H=Q_{8}C_{3}$ and since
$Q_{8}$ acts trivially on $O(G)$, we conclude that $G=O(G)\times H$. So $(3)$
implies $(2)$. ∎

For convenience we have divided into cases depending on whether or not $9$
divides the order of $G$. Now we will see another more unified approach. As
before let $Q_{8}$ be the unique $2$-Sylow subgroup. Let $S_{3}$ be a
$3$-Sylow subgroup of $G$. We note that $Q_{8}S_{3}$ maps onto $G/O(G)$, which
implies that $S_{3}$ is not in the centralizer of $Q_{8}$. So $S_{3}$ must act
on $Q_{8}$ by sending a generator to an automorphism of $Q_{8}$ of order 3.

Recall that $O(G)$ acts trivially on $Q_{8}$. Consider
$M=O(G)S_{3}=O(G)\rtimes S_{3}$. Note that the image of $M$ in
${\mathrm{Aut}}(Q_{8})$ has order 3. Also note that

$G=Q_{8}M=Q_{8}\rtimes M.$

######  Proposition 118.

Let $G$ be a Sylow-cycloidal group of binary tetrahedral type, and let $Q$ be
its unique $2$-Sylow subgroup of of $G$, which is isomorphic to the quaternion
group with 8 elements. Then there is a complement $M$ to $Q$ in $G$ which is a
Sylow-cycloidal group of odd order:

$G=QM=Q\rtimes M.$

Here $M\to{\mathrm{Aut}}Q$ has image of order $3$. Moreover, $G$ is freely
representable if and only if $M$ is freely representable.

###### Proof.

The only thing left to show is that if $M$ is freely representable, then $G$
is freely representable. Since $M$ is freely representable, then the same is
true of $O(G)$ since it is isomorphic to a subgroup of $M$. If $9$ divides the
order of $G$ then $G$ must be freely representable by Theorem 116. So from now
on we assume that $9$ does not divide the order of $G$.

Since $M$ is freely representable, it has a unique subgroup $C_{3}$ of order
$3$ (Theorem 81). Since $O(G)$ and $C_{3}$ are normal in $M$ we have
$M=O(G)\times C_{3}$ and $C_{3}$ acts trivially on $O(G)$. Note that $Q$ acts
trivially on $O(G)$, so $H=QC_{3}$ acts trivially on $O(G)$. This means that
$G=O(G)\times H$. However, $H$ is isomorphic to $G/O(G)$. Thus $G\cong
O(G)\times 2T$. So $G$ is freely representable by Theorem 117. ∎

We can strengthen the above:

######  Proposition 119.

Let $G$ be a finite group and let $Q$ be a $2$-Sylow subgroup of $G$. Then $G$
is a Sylow-cycloidal group of binary tetrahedral type if and only if $Q$ is a
normal quaternionic subgroup of $G$ and there exists a non-normal Sylow-cyclic
subgroup $M$ of index $|S|$ in $G$. In this case $G\cong Q\rtimes M$ where $Q$
is a quaternion group with 8 elements and where the action map
$M\to{\mathrm{Aut}}Q$ has image of size 3. Additionally, in this case $G$ is
freely representable if and only if $M$ is freely representable.

###### Proof.

In light of the previous proposition we just need to show that $G$ is a Sylow-
cycloidal group of binary tetrahedral type under the assumption that $Q$ is a
normal quaternionic group in $G$ and there exists a non-normal Sylow-cyclic
subgroup $M$ of index $|S|$ in $G$.

Under these assumption, every Sylow-subgroup of $M$ of odd order is actually a
Sylow-subgroup of $G$ since $[G:M]$ is even, and every Sylow-subgroup of $G$
is conjugate to a Sylow-subgroup of $M$ by the Sylow theorems. Thus every
Sylow-subgroup of $G$ of odd order is cyclic. Note also that $M$ is isomorphic
to $G/Q$ since $M$ is of odd order. Thus $G/Q$ and $Q$ are solvable, and so
$G$ is solvable.

Thus $G$ is a solvable Sylow-cycloidal group. In addition $O(G)$ is a proper
subgroup of $M$ since $O(G)$ is normal. So $G/O(G)$ is not a $2$-group. Thus
$G$ is not Sylow-cyclic, and is not of quaternion type. So $G$ is either of
binary tetrahedral or binary octahedral type. Note that the image of $Q$ in
$G/O(G)$ is a $2$-Sylow subgroup of $G/O(G)$ that is normal. This rules out
the binary octahedral type (since the existence of such a normal $2$-Sylow
subgroup in $2O$ gives a unique $2$-Sylow subgroup in its quotient $S_{4}$,
contradicting the fact that two-cycles of $S_{4}$ generates $S_{4}$). Thus $G$
is a Sylow-cycloidal group of binary tetrahedral type. ∎

###  5 Case 3: Binary Octahedral Type

We start with some basic observations about key subgroups of this type of
Sylow-cycloidal group:

######  Proposition 120.

Let $G$ be a Sylow-cycloidal group of binary octahedral type. Then every
$2$-Sylow subgroup of $G$ is a generalized quaternion group of order 16, and
these $2$-Sylow subgroups are not normal in $G$. In addition $G$ contains a
unique subgroup $H$ of index $2$, and this subgroup is of binary tetrahedral
type. Moreover, $G$ contains a unique quaternion subgroup $Q$ of order 8, and
this group $Q$ is contained in $H$. Finally, $G$ contains exactly four
subgroups of index $16$; these subgroups are conjugate subgroups of $H$ hence
are conjugate in $G$; these subgroups are not normal in $H$ hence are not
normal in $G$; these subgroups each contain $O(G)$ as a subgroup of index 3;
these subgroups are Sylow-cyclic groups of odd order; and these subgroups are
maximal among subgroups of odd order.

###### Proof.

Every $2$-Sylow subgroup $S$ of $G$ is isomorphic to a $2$-Sylow subgroup of
$G/O(G)\cong 2O$, so is a generalized quaternion group of order 16
(Proposition 105). The image $\overline{S}$ in $G/O(G)$ of a $2$-Sylow
subgroup $S$ contains the unique element of $G/O(G)$ of order 2, so these
Sylow subgroups $\overline{S}$ correspond to Sylow subgroups of the quotient
$O\cong S_{4}$. But Sylow subgroups of $S_{4}$ are not normal (a normal
$2$-Sylow subgroup $N$ in $S_{4}$ would have to contain all two cycles since
$S_{4}/N$ has order 3, but the collection of two cycles generate $S_{4}$).
Thus such an $\overline{S}$ is not normal in $G/O(G)$, and so $S$ cannot be
normal in $G$.

Let $H$ be the subgroup of $G$ containing $O(G)$ such that $H/O(G)$
corresponds to the binary tetrahedral subgroup of $G/O(G)$. Clearly
$O(G)\subseteq O(H)$, but since $H$ is normal in $G$ and since $O(H)$ is
characteristic in $H$, it follows that $O(H)$ is normal in $G$, and so
$O(G)=O(H)$. Hence $H/O(H)$ is a binary tetrahedral group.

Suppose $L$ is any subgroup of $G$ of index 2. Thus $L$ is normal in $G$.
Since the quotient $G/L$ has two elements and $O(G)$ has odd order, $O(G)$ is
contained in $L$. Note that $L/O(G)$ has index 2 in $G/O(G)$, so
$L/O(G)=H/O(G)$ (Proposition 104). This gives us $L=H$, and so $H$ is the
unique subgroup of index 2 in $G$.

By Proposition 114, $H$ has a unique subgroup $Q_{H}$ of order $8$ and this
group is a quaternion group. This group is characteristic in $H$ and so is
normal in $G$. By the Sylow theorems, $Q_{H}$ is contained in some $2$-Sylow
subgroup of $G$, and hence in all since $Q_{H}$ is normal (and all $2$-Sylow
subgroups are conjugate). Every quaternion subgroup $Q$ of $G$ of order $8$ is
contained in some $2$-Sylow subgroup $S$ of $G$ by the Sylow theorems, so both
$Q_{H}$ and $Q$ are subgroups of $S$. Thus $Q=Q_{H}$ since every general
quaternion group contains a unique subgroup isomorphic to the quaternion group
with 8 elements. So $Q_{H}$ is the unique subgroup $Q_{H}$ isomorphic to the
quaternion group with 8 elements.

Each subgroup of order $3$ in $G/O(G)$ is uniquely of the form $M/O(G)$ where
$M$ is a subgroup of $G$ containing $O(G)$. Each such $M$ is of index $16$ in
$G$ and contains $O(G)$ as a subgroup of index 3. Since each such $M$ is of
odd order, $M$ is a Sylow-cyclic group. Having index $16$ in $G$, each such
$M$ must be maximal among subgroups of odd order in $M$. Since $M\neq O(G)$
this means $M$ cannot be normal in $G$ since $O(G)$ is the maximal subgroup of
odd order.

Next we argue that each subgroup of index 16 in $G$ arises in this way.
Suppose $M$ has index $16$ in $G$. Then $O(G)M/O(G)$ is isomorphic to
$M/(M\cap O(G))$ which has odd order. Thus $O(G)M$ has odd order. But since
$M$ has index 16 in $G$, there is no strictly larger subgroup of odd order. So
$M=O(G)M$, and so $O(G)$ is contained in $M$. In particular, such a group
corresponds to a subgroup $M/O(G)$ of $G/O(G)$ of index 16 and order 3.

By Proposition 104, there are $4$ subgroups of $G/O(G)$ of order 3, and they
are all contained in $H/O(G)$. They constitute the $3$-Sylow subgroups of
$H/O(G)$ hence are conjugate in $H/O(G)$ by the Sylow theorems, and thus
cannot be normal in $H/O(G)$. This means that there are 4 subgroups of $G$ of
index 16, they are all contained in $H$, they are conjugate in $H$, and cannot
be normal in $H$. ∎

In particular, there is a unique subgroup of order $2$ in any group of this
type:

######  Proposition 121.

Let $G$ be a Sylow-cycloidal group of binary octahedral type. Then $G$ has a
unique element of order $2$.

###### Proof.

By the above proposition, we have a normal subgroup of order 8 in $G$, and
this group has a unique subgroup of order 2. Thus we have a normal subgroup of
order 2 in G. The result follows from Lemma 49. ∎

The main theorem about freely representable groups of this type is as follows:

######  Theorem 122.

Let $G$ be a Sylow-cycloidal group of binary octahedral type. Let $H$ be the
unique subgroup of $G$ of index $2$, which is of binary tetrahedral type. Let
$M$ be any of the four subgroups of $G$ of index 16, which is Sylow-cyclic
subgroup of $H$ of odd order. Then the following are equivalent

  1. 1.

$G$ is freely representable.

  2. 2.

$H$ is freely representable.

  3. 3.

$M$ is freely representable.

Furthermore, if $9$ divides the order of $G$ then if $O(G)$ is freely
representable, then so is $G$.

###### Proof.

The implication $(1)\implies(2)\implies(3)$ is clear since a subgroup of a
freely representable group is freely representable (Proposition 13). So we
just need to show that $(3)\implies(1)$. In fact we will proceed by showing
$(3)\implies(2)\implies(1)$.

Suppose that $M$ is freely representable. By Proposition 118, $H$ is also
freely representable. Since $G/H$ has order 2, any subgroup of $G$ odd prime
order $p$ is a subgroup of $H$. Also, by the previous proposition, $G$ has a
unique subgroup of order $2$ and this is a subgroup of $H$ since $H$ has even
order. By Corollary 80 we conclude that $G$ is freely representable.

Now suppose $9$ divides the order of $G$ and that $O(G)$ is freely
representable. Note that $O(H)$ is characteristic is $H$, and $H$ is normal in
$G$, so $O(H)$ is normal in $G$. This implies that $O(H)\subseteq O(G)$ so
that $O(H)$ is freely representable. Thus $H$ is freely representable by
Theorem 116, which as we have seen implies that $G$ is freely representable. ∎

###  6 Final Observations for the Solvable Case

We make some observations about solvable Sylow-cycloidal groups in general.

######  Proposition 123.

Let $G$ be a solvable Sylow-cycloidal group. If $G$ is not a Sylow-cyclic
group, then $G$ always has a unique element of order 2. If $G$ is a Sylow-
cyclic group then $G$ has a unique element of order 2 if and only if $G$ has a
normal subgroup of order 2.

###### Proof.

This was proved for each type individually. See Lemma 49, Proposition 109,
Corollary 115, and Proposition 121. ∎

######  Proposition 124.

Let $G$ be a solvable Sylow-cycloidal group of order $2^{k}n$ where $n$ is odd
and where $k\geq 1$. Then $G$ has a Sylow-cyclic subgroup $M$ of order $n$
with the following property: $G$ is freely representable if and only if (1)
$M$ is freely representable and (2) $G$ has a unique element of order $2$. In
particular, if $G$ is not itself Sylow-cyclic then $G$ is freely representable
if and only if $M$ is freely representable.

###### Proof.

This was proved for each type individually. See Proposition 108, Proposition
111, Proposition 118, and Theorem 122. ∎

##  Chapter 9 The Non-Solvable Case: the Group
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$.

A very important family of Sylow-cycloidal groups is
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$. The first goal here is to show that
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ is Sylow-cycloidal for all odd primes
$p$. In some sense these, together with the Sylow-cycloidal groups we have
considered up to now, are all that are needed to form the most general Sylow-
cycloidal groups. Along the way we see some very striking results about the
cyclic (or equivalently the Abelian) subgroups of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$. These results leads naturally to the
classification of normal subgroups of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$
and nonsolvability results. We will also consider an amusing necessary
condition for these groups to be freely-representable: $p$ is a Fermat prime.
It turns out that $p=3$ or $5$ is a necessary and sufficient condition, but we
will not prove this here (at least not in this version of the document).
Interestingly, $p=3$ and $p=5$ are the two cases that occur as subgroups of
${\mathbb{H}}^{\times}$.

A few of the initial results can be proved for any finite field $F$, but we
will need to specialize to $F$ of odd prime order if we want Sylow-cycloidal
groups.

######  Proposition 125.

Let $F$ be a finite field of order $q$. Then $\mathrm{SL}_{2}(F)$ is a group
of order $(q-1)q(q+1)$. If $q$ is odd then there is a unique element of order
$2$ in $\mathrm{SL}_{2}(F)$.

###### Proof.

The first row of an invertible $2$-by-$2$ matrix is nonzero, so we can limit
our attention to $q^{2}-1$ candidates for the top row. For each of these
candidates, we form an invertible matrix if and only if we choose the second
row not to be in the span of the first row. This gives $q^{2}-q$ choices for
the second row for each given top row. So there are

$(q^{2}-1)(q^{2}-q)=(q-1)^{2}q(q+1)$

elements of $\mathrm{GL}_{2}(F)$. The determinant homomorphism
$\mathrm{GL}_{2}(F)\to F^{\times}$ is surjective since even the invertible
diagonal matrices map onto $F^{\times}$. So the kernel of this map, which is
$\mathrm{SL}_{2}(F)$, has order $(q-1)q(q+1)$.

Now we assume $q$ is odd. Every element $\alpha$ of order 2 in
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$ has has an eigenvalue $1$ or $-1$ since
$(\alpha-I)(\alpha+I)=0$. Any element of $\mathrm{SL}_{2}({\mathbb{F}}_{p})$
with eigenvalue $\pm 1$ has the following form (with respect to some basis):

$\alpha=\pm\begin{pmatrix}1&a\\\
0&1\end{pmatrix},\qquad\alpha^{2}=\begin{pmatrix}1&2a\\\ 0&1\end{pmatrix}.$

We conclude that the only element of order 2 in
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$ is $-I$. ∎

Next we classify elements of order greater than 2 by the number of eigenvalues
in $F$, starting with two and working down to zero eigenvalues.

######  Lemma 126.

Let $F$ be a finite field of order $q$ and let $\alpha\in\mathrm{SL}_{2}(F)$
be an element with two distinct eigenvalues in $F$. Then $\alpha$ is contained
in a cyclic subgroup $C$ of $\mathrm{SL}_{2}(F)$ of order $q-1$ where $C$ has
the property that that there is a basis such that every element of $C$ is a
diagonal matrix. Furthermore, the existence of such an $\alpha$ implies that
$q>3$.

###### Proof.

Fix a basis of eigenvectors for $A$. Then $A$ is contained in the subgroup $C$
consisting of matrices of the form

$\begin{pmatrix}a&0\\\ 0&a^{-1}\end{pmatrix}$

with respect to the chosen basis, where $a\in F^{\times}$. Since $F^{\times}$
is cyclic, this group is a cyclic group of order $q-1$.

Since the product of the eigenvalues of $\alpha$ is $1$ and since they are
distinct, we must have $q>3$. ∎

######  Lemma 127.

Let $F$ be a finite field of order $q$ and characteristic $p$. Let $v\in
F^{2}$ be a nonzero vector. Then the set of elements $D_{v}$ of
$\mathrm{SL}_{2}(F)$ with exactly one eigenvalue and with eigenvector $v$ is a
subgroup of $\mathrm{SL}_{2}(F)$ isomorphic to $\\{\pm 1\\}\times F$ where $F$
is the additive group of $F$ (of size $q$). In particular $D_{v}$ is Abelian
and each element of $F$ has order divisible by $2p$ if $q$ is odd, and
divisible by $p=2$ if $q$ is even.

Let $\alpha\in\mathrm{SL}_{2}(F)$ be an element with exactly one eigenvalue in
$F$. If $q$ is odd then $\alpha$ is a contained in a cyclic subgroup $C$ of
$\mathrm{SL}_{2}(F)$ of order $2p$. If $q$ is even then $\alpha$ is contained
in a cyclic subgroup $C$ of $\mathrm{SL}_{2}(F)$ of order $p=2$.

###### Proof.

Choose a basis whose first element is $v$. Then $D_{v}$ consists of the
matrices whose representation with respect to this basis is of the form

$\pm\begin{pmatrix}1&a\\\ 0&1\end{pmatrix}$

for some $a\in F$. Observe that

$\begin{pmatrix}1&a\\\ 0&1\end{pmatrix}\begin{pmatrix}1&b\\\
0&1\end{pmatrix}=\begin{pmatrix}1&a+b\\\ 0&1\end{pmatrix},$

and in fact that $D_{v}$ is isomorphic to $\\{\pm 1\\}\times F$. In
particular, every element of $D_{v}$ is contained in a cyclic subgroup of
order $2p$ if $q$ is odd, and a subgroup of order $p=2$ if $q$ is even. ∎

######  Lemma 128.

Let $F$ be a finite field of order $q$ and let $\alpha\in\mathrm{SL}_{2}(F)$
be an element with no eigenvalues in $F$. Then $\alpha$ is contained in a
cyclic subgroup of $\mathrm{SL}_{2}(F)$ of order $q+1$. More specifically, let
$M_{2}(F)$ be the ring of $2$-by-$2$ matrices with entries in $F$, where we
view $F$ as a subring via the diagonal embedding. Then the subring
$E=F[\alpha]$ of $M_{2}(F)$ generated by $F$ and $\alpha$ is a field. This
field $E$ has the following properties:

  * •

$E$ has size $q^{2}$.

  * •

The group $K={\mathrm{SL}}_{2}(F)\cap E^{\times}$ is a cyclic group of order
$q+1$ that contains $\alpha$.

  * •

The determinant homomorphism $E^{\times}\to F^{\times}$ is the map $x\mapsto
x^{q+1}$, and the kernel of this map is $K={\mathrm{SL}}_{2}(F)\cap
E^{\times}$.

  * •

The Galois group of $E$ over $F$ has two elements. Its nontrivial element
$\sigma$ is the automorphism $x\mapsto x^{q}$. For all $x\in
K={\mathrm{SL}}_{2}(F)\cap E^{\times}$ we have $\sigma x=x^{-1}$.

###### Proof.

Using the ring homomorphism $F[X]\to M_{2}(F)$ sending $X$ to $\alpha$, we see
that $F[\alpha]$ is isomorphic to $F[X]/\left<f\right>$ where $F[X]$ is the
polynomial ring in one-variable and $f$ is the minimal polynomial of $\alpha$
in $F[X]$. By the Cayley-Hamilton theorem, $f$ divides the characteristic
polynomial of $\alpha$, so in this case $f$ must be an irreducible quadratic
polynomial since the characteristic polynomial of $\alpha$ has no roots in
$F$. This implies that $F[X]/\left<f\right>$ is a quadratic field extension of
$F$. In particular, $E=F[\alpha]$ is a field of size $q^{2}$. Note that
$E^{\times}$ is cyclic of order $q^{2}-1$ (Corollary 92).

Because $F^{\times}$ is the subgroup of $E^{\times}$ of size $q-1$, we have
that that $\beta\in E^{\times}$ is in $F^{\times}$ if and only if
$\beta^{q-1}=1$. This implies that $\beta\in E$ is in $F$ if and only if
$\beta^{q}=\beta$.

Now let $\beta$ be a generator of the cyclic group $E^{\times}$. In $E[X]$ we
have the polynomial

$(X-\beta)(X-\beta^{q})=X^{2}-(\beta+\beta^{q})X+\beta^{q+1}$

Observe that

$(\beta+\beta^{q})^{q}=\beta^{q}+\beta^{q^{2}}=\beta^{q}+\beta$

and that

$(\beta^{q+1})^{q}=\beta^{q^{2}}\beta^{q}=\beta^{1+q}.$

These follow since $\beta^{q^{2}-1}=1$, so $\beta^{q^{2}}=\beta$. Also $q$ is
a power of the characteristic $p$ of $E$ and in fields of characteristic $p$
we have the identity $(a+b)^{p}=a^{p}+b^{p}$, so $(a+b)^{q}=a^{q}+b^{q}$ for
all $a,b\in E$. We conclude that $X^{2}-(\beta+\beta^{q})X+\beta^{q+1}$ lies
in $F[X]$ and so must be the minimal polynomial of $\beta$ in $F[X]$ (since
$\beta$ is not in $F$). By the Cauchy-Hamilton theorem, it is the
characteristic polynomial of $\beta$. In particular the determinant of $\beta$
is $\beta^{q+1}$. Note also that we have established that $x\mapsto x^{q}$ is
an automorphism $\sigma$ of the field $E$, and that it fixes $F$.

Consider the determinant homomorphism $E^{\times}\to F^{\times}$. This sends
the generator $\beta$ to $\beta^{q+1}$, so sends any element in $E^{\times}$
to its $q+1$ power. In particular the kernel $K$ must be the cyclic subgroup
of $E^{\times}$ of size $q+1$ since $q+1$ divides $q^{2}-1$. Note also that
$K=E^{\times}\cap{\mathrm{SL}}_{2}(F)$ since it is in the kernel of the
determinant map. So $\alpha\in K$.

Also note that any automorphism of $E$ fixing $F$ is determined by its action
on the generator $\beta$ of $E^{\times}$, and that $\beta$ must map to a root
of $(X-\beta)(X-\beta^{q})$. Thus there are only two elements of the Galois
group of $E$ over $F$ (i.e., the automorphisms of $E$ fixing $F$): the
identity $x\mapsto x$ and $x\mapsto x^{q}$. Let $\sigma$ be the map $x\mapsto
x^{q}$. Note that if $x\in K$ then $x^{q+1}=1$ so $x^{q}=x^{-1}$. Thus the
restriction of $\sigma$ to $K$ is the map $x\mapsto x^{-1}$. ∎

We can combine these three lemmas:

######  Lemma 129.

Let $F$ be a finite field of order $q$ and characteristic $p$. Let
$\alpha\in\mathrm{SL}_{2}(F)$ be an element not equal to $1$ or $-1$. Then the
following hold:

  * •

The element $\alpha$ has two distinct eigenvalues in $F$ if and only if
$\alpha$ has order dividing $q-1$.

  * •

The element $\alpha$ has exactly one eigenvalue in $F$ if and only if $\alpha$
has order dividing $2p$.

  * •

The element $\alpha$ has no eigenvalues in $F$ if and only if $\alpha$ has
order dividing $q+1$.

Furthermore $\alpha$ can only have two distinct eigenvalues if $q>3$.

###### Proof.

One direction of each implication follows directly from Lemmas 126, 127, and
128. To see the converse, observe that the GCD of any two distinct elements of
$\\{q-1,2p,q+1\\}$ is $2$ is $q$ is odd, and is $1$ is $q$ is even. Now if $q$
is odd then $-1$ is the only element of order $2$ (Proposition 125). Since
$\alpha$ is not $1$ or $-1$, we conclude that the order of $\alpha$ cannot
divide two of $\\{q-1,2p,q+1\\}$. From this the converse follows. ∎

A _maximal_ cyclic subgroup of a finite group $G$ is defined to be a cyclic
subgroup of $G$ that is not contained in a cyclic group of $G$ of larger
order. Of course every cyclic subgroup is contained in an maximal cyclic group
(since $G$ is finite), but a given cyclic subgroup might be contained in
several maximal cyclic groups in a general group $G$. For
$G={\mathrm{SL}}_{2}(F)$ we can establish uniqueness.

######  Corollary 130.

Let $F$ be a finite field of order $q$ and characteristic $p$ and let $C$ be a
cyclic subgroup of ${\mathrm{SL}}_{2}(F)$.

  * •

If $q>3$ is odd then $C$ is a maximal cyclic subgroup if and only if it has
order $q-1$, $2p$, or $q+1$.

  * •

If $q=3$ then $C$ is a maximal cyclic subgroup if and only if it has order
$2p=6$ or $q+1=4$.

  * •

If $q>2$ is even then $C$ is a maximal cyclic subgroup if and only if it has
order $q-1$, $2$, or $q+1$.

  * •

If $q=2$ then $C$ is a maximal cyclic subgroup if and only if it has order $2$
or $3$.

###### Proof.

Let $\alpha$ be a generator of $C$. Since $C$ is a maximal cyclic group, it
has the listed orders by Lemmas 126, 127, and 128 depending on the number of
eigenvalues of $\alpha$ in $F$.

Conversely, suppose $q>3$ is odd and suppose $C$ has order exactly $q-1$,
$2p$, or $q+1$. Let $D$ be a maximal cyclic subgroup containing $C$. Then $D$
also has order $q-1$, $2p$, or $q+1$ by the above. Note that (1) the order of
$C$ divides the order of $D$, (2) the order of $C$ is at least $3$ (since
$q>3$), and (3) the GCD of any two of $\\{q-1,2p,q+1\\}$ is $2$. So it is
impossible for $C$ and $D$ to have different orders. So $C=D$. A similar but
modified argument works for $q=3$, or $q>2$ even, or $q=2$. ∎

We are now ready for the first main theorem.

######  Theorem 131.

Let $p$ be an odd prime. Then ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ is a
Sylow-cycloidal group whose $2$-Sylow subgroups are not cyclic.

###### Proof.

Let $q$ be an odd prime dividing the order $(p-1)p(p+1)$ of the group
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$ and let $q^{k}$ be the maximal power that
$q$ divides $(p-1)p(p+1)$. By Cauchy’s theorem, there is an element $\alpha$
of order $q$. Let $C$ be a maximal cyclic subgroup
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$ containing $\alpha$. By the above
corollary, $C$ has order $p-1$ or $2p$ or $p+1$. Since $q\geq 3$, we have that
$q$ and hence $q^{k}$ divides exactly one of $p-1$ or $p$ or $p+1$. Since $q$
divides $|C|$ this means that if $|C|=p-1$ or $p+1$ then $q^{k}$ divides
$|C|$. If $|C|=2p$ then $q=p$ (since $q$ is odd) and so $q^{k}$ divides $|C|$
as well (and $k=1$ in this case). In any case, $C$ has a cyclic subgroup of
order $q^{k}$. Since all $q$-Sylow subgroups are conjugate, we have
established that all $q$-Sylow subgroups are cyclic.

Since ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ has a unique element of order $2$
(Proposition 125), each $2$-Sylow subgroup of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ has a unique element of order $2$ (by
Cauchy’s theorem). Thus each $2$-Sylow subgroup $S$ of
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$ is either cyclic or quaternionic
(Corollary 37). So ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ is a Sylow-cycloidal
group.

Suppose $S$ is a cyclic $2$-subgroup of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$,
and let $C$ be a maximal cyclic subgroup containing $S$. By the above
corollary, $S$ has order dividing $|C|$ which is either $p-1$ or $2p$ or
$p-1$. Since $2$ divides both $p-1$ and $p+1$, the largest power of $2$ diving
$|C|$ is less than the largest power of $2$ dividing the order $(p-1)p(p+1)$
of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$. Thus $S$ is not a 2-Sylow subgroup.
∎

Here is a partial converse.

######  Proposition 132.

Suppose that $F$ is a finite field. If ${\mathrm{SL}}_{2}(F)$ is a Sylow-
cycloidal group then $F={\mathbb{F}}_{p}$ for some prime $p$.

###### Proof.

By Lemma 127 there is a subgroup $D$ of ${\mathrm{SL}}_{2}(F)$ isomorphic to
the additive group $F$. If ${\mathrm{SL}}_{2}(F)$ is a Sylow-cycloidal group
then the Abelian group $D\cong F$ must be cyclic. This can only happen if
$F={\mathbb{F}}_{p}$ where $p$ is the characteristic of $F$. ∎

######  Remark.

The case ${\mathrm{SL}}_{2}({\mathbb{F}}_{2})$ is special. It has order $6$,
and so is Sylow-cyclic since it is of prime free order. As we have seen, all
its cyclic subgroups are of order $2$ or $3$, so it is not cyclic and so must
be dihedral. Note that ${\mathrm{SL}}_{2}({\mathbb{F}}_{2})$ fails to have a
unique element of order $2$.

The next major result is that all cyclic subgroups of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ of the same order are conjugate when $p$
is a prime. It is a bit easier to show that such groups are conjugate in
$\mathrm{GL}_{p}({\mathbb{F}}_{p})$, but the following two lemmas will give us
tools to achieve conjugacy in ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ instead of
$\mathrm{GL}_{p}({\mathbb{F}}_{p})$.

######  Lemma 133.

Let $\alpha\in\mathrm{GL}_{2}(F)$ where $F$ is a finite field of order $q$ and
let $d\in F^{\times}$. If $\alpha$ has two distinct eigenvalues in $F$ or no
eigenvalues in $F$ then there is an element $\beta\in\mathrm{GL}_{2}(F)$ such
that $\beta\alpha\beta^{-1}=\alpha$ and $\det\beta=d$.

###### Proof.

Suppose $\alpha$ has two eigenvalues in $F$ and let $v_{1},v_{2}\in F^{2}$ be
a basis of eigenvectors. Then let $\beta$ be such that its associated linear
tranformation maps $v_{1}\mapsto dv_{1}$ and $v_{2}\mapsto v_{2}$. Then
$\beta$ clearly works.

Suppose that $\alpha$ has no eigenvalues in $F$. By Lemma 128, $\alpha\in
E^{\times}$ where $E^{\times}$ is a cyclic subgroup of $\mathrm{GL}_{2}(F)$ of
order $q^{2}-1$ (in fact $E^{\times}$ is the multiplicative group of a field);
furthermore, the determinant map on $E^{\times}\to F^{\times}$ is given by
$x\mapsto x^{q+1}$ and so is surjective since $F^{\times}$ has order $q-1$.
Now just let $\beta\in E^{\times}$ be an element of determinant $d$. ∎

######  Lemma 134.

Let $p$ be a prime and let $C$ be a cyclic subgroup of
${\mathrm{SL}}({\mathbb{F}}_{p})$. Let $d\in{\mathbb{F}}_{p}^{\times}$. Then
there is an there is an element $\beta\in\mathrm{GL}_{2}(F)$ such that $\beta
C\beta^{-1}=C$ and $\det\beta=d$.

###### Proof.

Let $\alpha$ be a generator of $C$. If $\alpha$ has two distinct eigenvalues
or no eigenvalues then the result follows from the previous lemma. So suppose
that $\alpha$ has exactly one eigenvalue in $F$ and let $v_{1}$ be an
eigenvector. Let $v_{2}$ be such that $v_{1},v_{2}$ form a basis for $F^{2}$.
The representation for $\alpha$ in this basis is of the form

$\begin{pmatrix}e&a\\\ 0&e\end{pmatrix}$

where $e$ is $1$ or $-1$. So let $\beta$ be an element with the following
matrix representation (for this same basis $v_{1},v_{2}$):

$\begin{pmatrix}d&0\\\ 0&1\end{pmatrix}$

Note that

$\begin{pmatrix}d&0\\\ 0&1\end{pmatrix}\begin{pmatrix}e&a\\\
0&e\end{pmatrix}\begin{pmatrix}d^{-1}&0\\\
0&1\end{pmatrix}=\begin{pmatrix}e&da\\\ 0&e\end{pmatrix}=\begin{pmatrix}e&a\\\
0&e\end{pmatrix}^{m}$

where $m$ is an odd positive integer such that $m\equiv d\pmod{p}$. ∎

As mentioned above, our next goal is to establish that cyclic subgroups
(equivalently, Abelian subgroups) of the same order of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ are conjugate. We start with cyclic
groups of order $p-1$, followed by order $p$, then order $p+1$, ending with
general order.

######  Lemma 135.

Let $F$ be a finite field of order $q$. The vector space $F^{2}$ has $q+1$
distinct one-dimensional subspaces. Let $L_{1}$ and $L_{2}$ be two distinct
one-dimensional subspaces of $F^{2}$. Then the set elements of
${\mathrm{SL}}_{2}(F)$ with a basis of eigenvectors in $L_{1}\cup L_{2}$ forms
a cyclic subgroup of ${\mathrm{SL}}_{2}(F)$ of order $q-1$.

Now assume $q>3$. Then all cyclic subgroups of ${\mathrm{SL}}_{2}(F)$ of order
$q-1$ arise in this way. There are $\frac{1}{2}q(q+1)$ cyclic subgroups of
order $q-1$ and they are conjugate. Two distinct cyclic subgroups of order
$q-1$ have intersection $\\{\pm 1\\}$.

###### Proof.

Counting the number of one-dimensional subspaces is straightforward.

Let $v_{1},v_{2}$ be a basis for $F^{2}$ such that $v_{1}\in L_{1}$ and
$v_{2}\in L_{2}$. Then $\alpha\in{\mathrm{SL}}_{2}(F)$ has a basis of
eigenvectors in $L_{1}\cup L_{2}$ if and only if it has form

$\alpha=\begin{pmatrix}a&0\\\ 0&a^{-1}\end{pmatrix}$

with respect to this basis, where $a\in F^{\times}$. So the set of such
matrices is a cyclic subgroup of order $q-1$.

Assume $q>3$ for the remainder of the proof.

Let $C$ be a cyclic subgroup of ${\mathrm{SL}}_{2}(F)$ of order $q-1$, and let
$\alpha$ be a generator of $C$. So $\alpha$ is not $1$ or $-1$ since $q-1>2$.
By Lemma 129, $\alpha$ has two distinct eigenvalues, and its eigenvectors
determine two distinct one-dimensional subspaces $L_{1}$ and $L_{2}$ of
$F^{2}$. So $\alpha$ is in a cyclic group of order $q-1$ of the form described
above. Hence $C$ has the desired form.

Suppose $C$ and $C^{\prime}$ are cyclic subgroup of ${\mathrm{SL}}_{2}(F)$ of
order $q-1$. Suppose $C$ is defined using $L_{1},L_{2}$ and $C^{\prime}$ is
define using $L_{1}^{\prime},L_{2}^{\prime}$. Suppose $g\in C\cap C^{\prime}$
is not $\pm 1$. Then $g$ has distinct eigenvalues (Lemma 129). Let $v_{1}$ be
an eigenvector of $g$. Then $v_{1}\in L_{i}$ and $v_{1}\in L^{\prime}_{j}$ for
some $i,j$, so $L_{i}=L^{\prime}_{j}$ since these are one-dimensional. After
renumbering we can assume $L_{1}=L^{\prime}_{1}$. Let $v_{2}$ be an
eigenvector of $g$ not in the span of $v_{1}$. Then $v_{2}$ is in $L_{2}$ and
$L_{2}^{\prime}$ so $L_{2}=L^{\prime}_{2}$. Thus $C=C^{\prime}$. In other
words, distinct cyclic subgroup of ${\mathrm{SL}}_{2}(F)$ of order $q-1$
intersect in $\\{\pm 1\\}$. (Note that $-1\in C$ and $-1\in C^{\prime}$ by
Proposition 125 if $q$ is odd, and trivially if $q$ is even).

So different choices of $\\{L_{1},L_{2}\\}$ will produce difference cyclic
subgroups of order $q-1$ since $q-1>2$. So there are $\frac{1}{2}q(q+1)$ such
cyclic subgroups.

Suppose $C$ and $C^{\prime}$ are cyclic subgroup of ${\mathrm{SL}}_{2}(F)$ of
order $q-1$. Suppose $C$ is defined using $L_{1},L_{2}$ and $C^{\prime}$ is
define using $L_{1}^{\prime},L_{2}^{\prime}$. Let $\alpha$ be a generator for
$C$. Let $\beta_{1}\in\mathrm{GL}_{2}(F)$ give a linear transformation mapping
$L_{1}$ to $L_{1}^{\prime}$ and $L_{2}$ to $L_{2}^{\prime}$. Let
$\beta_{2}\in\mathrm{GL}_{2}(F)$ be such that
$\beta_{2}\alpha\beta^{-1}_{2}=\alpha$ and $\det(\beta_{1}\beta_{2})=1$ (see
Lemma 133). Let $\beta=\beta_{1}\beta_{2}$. Observe that $C^{\prime}=\beta
C\beta^{-1}$. So $C$ and $C^{\prime}$ are conjugate. ∎

######  Lemma 136.

Let $p$ be an odd prime. The vector space ${\mathbb{F}}_{p}^{2}$ has $p+1$
distinct one-dimensional subspaces. Let $L$ be a one-dimensional subspaces of
${\mathbb{F}}_{p}^{2}$. Then the set elements of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ with exactly one eigenvalue, and with an
eigenvector in $L$, forms a cyclic subgroup of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ of order $2p$.

All cyclic subgroups of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ of order $2p$
arise in this way. There are $p+1$ cyclic subgroups of order $2p$ and they are
conjugate. Two distinct cyclic subgroups of order $2p$ have intersection
$\\{\pm I\\}$.

###### Proof.

Counting the number of one-dimensional subspaces is straightforward.

Let $v_{1},v_{2}$ be a basis for ${\mathbb{F}}_{p}^{2}$ such that $v_{1}\in
L$. Then $\alpha\in{\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ has exactly one
eigenvalue and has an eigenvector in $L$ if and only if can be written as

$\alpha=\pm\begin{pmatrix}1&a\\\ 0&1\end{pmatrix}$

with respect to this basis, where $a\in{\mathbb{F}}_{p}$. Observe that the set
of such matrices forms a cyclic group isomorphic to $\\{\pm
1\\}\times{\mathbb{F}}_{p}$ where ${\mathbb{F}}_{p}$ here is the additive
group.

Let $C$ be a cyclic subgroup of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ of order
$2p$, and let $\alpha$ be a generator of $C$. Then $\alpha$ has exactly one
eigenvalue by Lemma 129. If $L$ is an eigenspace for $\alpha$, then as above
$\alpha$ is contained in a group of order $2p$ of the given form. So $C$ is a
group of the desired form.

Suppose $C$ and $C^{\prime}$ are cyclic subgroup of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ of order $2p$ where $C$ is defined using
$L$ and $C^{\prime}$ is define using $L^{\prime}$. Suppose $g\in C\cap
C^{\prime}$ is not $\pm 1$. Then $g$ has a unique eigenvalue, and the
eigenspace associated to that eigenvalue has dimension one (since $g\neq\pm
1$). Thus $L=L^{\prime}$ and so $C=C^{\prime}$. In other words, distinct
cyclic subgroup of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ of order $2p$
intersect in $\\{\pm 1\\}$.

So different choices of $L$ will produce difference cyclic subgroups of order
$2p$. So there are $p+1$ such cyclic subgroups.

Suppose $C$ and $C^{\prime}$ are cyclic subgroup of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ of order $2p$. Suppose $C$ is defined
using $L$ and $C^{\prime}$ is define using $L^{\prime}$. Let
$\beta_{1}\in\mathrm{GL}_{2}({\mathbb{F}}_{p})$ represent a linear
transformation sending $L$ to $L^{\prime}$. Let
$\beta_{2}\in\mathrm{GL}_{2}(F)$ be such that $\beta_{2}C\beta^{-1}_{2}=C$ and
so that $\det(\beta_{1}\beta_{2})=1$ (see Lemma 134). Let
$\beta=\beta_{1}\beta_{2}$. Observe that $C^{\prime}=\beta C\beta^{-1}$. So
$C$ and $C^{\prime}$ are conjugate. ∎

######  Lemma 137.

Let $F$ be a finite field of order $q$ and let $M_{2}(F)$ be the ring of
$2$-by-$2$ matrices with entries in $F$. Then the following hold:

  * •

If $C$ is a cyclic subgroup of ${\mathrm{SL}}_{2}(F)$ of order $q+1$ then
there is a unique quadratic field extension $E$ of $F$ in $M_{2}(F)$
containing $C$, and $C$ is the unique subgroup of $E^{\times}$ of order $q+1$.

  * •

Two distinct cyclic subgroups of ${\mathrm{SL}}_{2}(F)$ of order $q+1$ are
conjugate in ${\mathrm{SL}}(F)$ and have intersection $\\{\pm 1\\}$.

  * •

If $C$ is a cyclic subgroup of ${\mathrm{SL}}_{2}(F)$ of order $q+1$ then
there is a $\gamma\in{\mathrm{SL}}_{2}(F)$ such that $x\mapsto\gamma
x\gamma^{-1}$ is an automorphism of the group $C$ sending any $x\in C$ to
$x^{-1}$.

###### Proof.

Let $\alpha$ be a generator of $C$. By Lemma 129, $\alpha$ cannot have
eigenvalues in $F$. So by Lemma 128, the subring $E=F[\alpha]$ of of
$M_{2}(F)$ is a field of order $q^{2}$, and $\alpha$ is contained in
$K={\mathrm{SL}}_{2}(F)\cap E^{\times}$, which is the unique subgroup of
$E^{\times}$ of order $q+1$. So in fact $C=K$. Since $E=F[\alpha]$ any
quadratic field extension $E$ of $F$ in $M_{2}(F)$ containing $C$ would
contain $\alpha$ and so $F[\alpha]$, and hence be equal to $E$. So $E$ is the
unique such field.

Suppose $C$ and $C^{\prime}$ are two subgroups of ${\mathrm{SL}}_{2}(F)$ size
$q+1$ and that $E$ and $E^{\prime}$ are the respective quadratic extensions of
$F$ containing $C$ and $C^{\prime}$. Suppose $\beta\in C\cap C^{\prime}$ is
not $\pm 1$. Then $\beta$ cannot be of the form $c1$ for some $c\in F$ since
the determinant of $\beta$ is $1$. Thus $\beta$ generates a proper extension
of $F$ contained in the intersection $E\cap E^{\prime}$. So
$E=E^{\prime}=F[\beta]$ since $[E:F]=[E:F]=2$ and $[E^{\prime}:F]\geq 2$. We
conclude that $C=C^{\prime}$. In other words, if $C$ and $C^{\prime}$ are
distinct subgroups of ${\mathrm{SL}}_{2}(F)$ of size $q+1$ then their
intersection is $\\{\pm 1\\}$.

Next fix a generator $g$ of $F^{\times}$. In other words, $g$ has order $q-1$.
Suppose $C$ is a cyclic subgroup of ${\mathrm{SL}}_{2}(F)$ of order $q+1$ and
let $E$ be the field extension of $F$ inside $M_{2}(F)$ of size $q^{2}$
containing $C$. Since $E^{\times}$ has order $q^{2}-1$, and since $q+1$ is
even, there is a subgroup of $E^{\times}$ of order $2(q-1)$. This implies that
$E^{\times}$ has an element $\beta$ such that $\beta^{2}=g$. Fix a nonzero
vector $v_{1}\in F^{2}$ and let $v_{2}=\beta v$ (viewing $v$ as a column
vector). By the Cayley-Hamilton theorem, the characteristic polynomial of
$\beta$ is $X^{2}-g$, so $\beta$ has no eigenvalues in $F$. Thus $v_{1},v_{2}$
must be a basis of $F^{2}$. Note also that $\beta v_{2}=gv_{1}$.

Suppose $C^{\prime}$ is another cyclic subgroup of ${\mathrm{SL}}_{2}(F)$ of
order $q+1$ and let $E^{\prime}$ be the field extension of $F$ inside
$M_{2}(F)$ of size $q^{2}$ containing $C^{\prime}$. Let
$\beta^{\prime}\in(E^{\prime})^{\times}$ be such that $v_{1}^{\prime}=v_{1}$
and $v^{\prime}_{2}=\beta^{\prime}v^{\prime}_{1}$ forms a basis with
$\beta^{\prime}v_{2}^{\prime}=gv_{1}^{\prime}$.

Let $\gamma_{1}\in\mathrm{GL}_{2}(F)$ be chosen so that $\gamma_{1}$ maps
$v_{1}=v_{1}^{\prime}$ to itself, and maps $v_{2}$ to $v_{2}^{\prime}$. Note
that $\gamma_{1}\beta\gamma_{1}^{-1}=\beta^{\prime}$ since both sides of this
equation map $v^{\prime}_{1}\mapsto v^{\prime}_{2}$ and $v^{\prime}_{2}\mapsto
gv^{\prime}_{1}$. Since $F[\beta]=E$ and $F[\beta^{\prime}]=E^{\prime}$ this
means that the map

$x\mapsto\gamma_{1}x\gamma_{1}^{-1}$

is an isomorphism $E\to E^{\prime}$ between fields. Hence it sends the unique
subgroup $C$ of $E^{\times}$ of order $q+1$ to the unique subgroup
$C^{\prime}$ of $(E^{\prime})^{\times}$ of order $q+1$.

In other words, $\gamma_{1}C\gamma_{1}^{-1}=C^{\prime}$. Let
$\gamma_{2}\in\mathrm{GL}_{2}(F)$ be chosen so that $\gamma_{1}\gamma_{2}$ has
determinant 1 and so that $\gamma_{2}C\gamma_{2}^{-1}=C$ (Lemma 133 applied to
a generator $\alpha$ of $C$ and to $d=\det\gamma_{1}^{-1}$). If
$\gamma=\gamma_{1}\gamma_{2}$ then $\gamma C\gamma^{-1}=C^{\prime}$ and
$\gamma\in{\mathrm{SL}}_{2}(F)$. So $C$ and $C^{\prime}$ are conjugate in
${\mathrm{SL}}_{2}(F)$.

A special case of this construction is where $C=C^{\prime}$ and where
$\beta^{\prime}=-\beta$ so that $\gamma_{1}\beta\gamma_{1}^{-1}=-\beta$.
Observe that $x\mapsto\gamma_{1}x\gamma_{1}^{-1}$ must then be a field
automorphism fixing $F$ (thought of as diagonal matrices). Since $F^{\times}$
has $C$ for its unique subgroup of order $q+1$, this automorphism acts on $C$.
Lemma 128 implies that it sends $x\in C$ to $x^{-1}$. So if
$\gamma=\gamma_{1}\gamma_{2}$ then $\gamma x\gamma^{-1}=x^{-1}$ for all $x\in
C$. ∎

These lemmas combine to yield the following:

######  Lemma 138.

Let $p$ be an odd prime, and let $C$ and $C^{\prime}$ be two distinct maximal
cyclic subgroups of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$. Then $C\cap
C^{\prime}=\\{\pm 1\\}$. Furthermore, if $C$ and $C^{\prime}$ have the same
order then $C$ and $C^{\prime}$ are conjugate subgroups of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$.

###### Proof.

By Corollary 130, the orders of $C$ and $C^{\prime}$ are in the set
$\\{p-1,2p,p+1\\}$. If $C$ and $C^{\prime}$ happen to have different orders
then $C\cap C^{\prime}$ has order dividing $2$ since the GCD of any two of
$\\{p-1,2p,p+1\\}$ is $2$. Since $C$ and $C^{\prime}$ are of even order they
both contained $-1$ (since $-1$ is the unique element of order 2). So $C\cap
C^{\prime}=\\{\pm 1\\}$.

Now we consider the case where $C$ and $C^{\prime}$ have the same order in
$\\{p-1,2p,p+1\\}$. Then $C$ and $C^{\prime}$ are conjugate and $C\cap
C^{\prime}=\\{\pm 1\\}$ by Lemma 135, Lemma 136, or Lemma 136 (depending on
the order of $C$). ∎

The above lemma yields the conjugacy results we want but only for _maximal_
cyclic subgroups of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$. The following lemma
and proposition make it possible to establish the result for cyclic subgroups
more generally.

######  Lemma 139.

Let $C$ be a cyclic subgroup of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ where
$p$ is an odd prime. If the order of $C$ is at least $3$ then there is a
unique maximal cyclic subgroup of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$
containing $C$.

###### Proof.

Suppose $D_{1}$ and $D_{2}$ are distinct maximal cyclic subgroups of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ containing $C$. By the above lemma
$D_{1}\cap D_{2}=\\{\pm 1\\}$, contradicting the assumption on the size of
$C$. ∎

######  Proposition 140.

Let $C$ be a cyclic subgroup of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ where
$p$ is a odd prime. If $C$ has order at least $3$ then the centralizer $Z(C)$
is the unique maximal cyclic subgroup of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$
containing $C$. So if $C$ is a maximal cyclic subgroup of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ then $C=Z(C)$.

###### Proof.

Let $D$ be the unique maximal cyclic subgroup of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ containing $C$ (see previous lemma).
Suppose that $h\in Z(C)$ and let $H$ be the subgroup generated by $h$. Observe
that $HC$ is an Abelian subgroup of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$.
Since ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ is Sylow-cycloidal (Theorem 131)
the group $HC$ must be cyclic and so is contained in a maximal cyclic subgroup
$D^{\prime}$. By the previous lemma $D=D^{\prime}$ since both contain $C$.
Thus $h\in D$. We have established that $Z(C)\subseteq D$. The other inclusion
is clear. ∎

We are ready for the second main theorem.

######  Theorem 141.

Let $p$ be an odd prime, and let $C$ and $C^{\prime}$ be two cyclic subgroups
of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ of the same order. Then $C$ and
$C^{\prime}$ are conjugate subgroups of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$.
If, in addition, $C\neq C^{\prime}$ then $C\cap C^{\prime}\subseteq\\{\pm
1\\}$.

###### Proof.

If $C$ and $C^{\prime}$ have order $1$ or $2$ then the result is clear since
$-1$ is the unique element of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ of order
$2$. So from now on assume that $C$ and $C^{\prime}$ have equal order at least
$3$. By the above proposition, $Z(C)$ and $Z(C^{\prime})$ are maximal cyclic
subgroups. By Corollary 130, the orders of $Z(C)$ and $Z(C^{\prime})$ are
restricted to the set $\\{p-1,2p,p+1\\}$. So the order of $C$ and $C^{\prime}$
must divide an element of the set $\\{p-1,2p,p+1\\}$, and in fact divides a
unique element of $\\{p-1,2p,p+1\\}$ since the GCD of any two distinct
elements is $2$. This means that $Z(C)$ and $Z(C^{\prime})$ have the same
order, namely the unique element of $\\{p-1,2p,p+1\\}$ that is a multiple of
the order of $C$.

Thus $Z(C)$ and $Z(C^{\prime})$ are conjugate subgroups by Lemma 138. Since
$C$ is the unique subgroup of $Z(C)$ of its order, and the same is true of
$C^{\prime}$ in $Z(C^{\prime})$, we conclude that $C$ and $C^{\prime}$ are
conjugate as well.

If, in addition, $C\neq C^{\prime}$ then $Z(C)$ cannot equal $Z(C^{\prime})$
since $C$ is the unique subgroup of $Z(C)$ of its order, and the same is true
of $C^{\prime}$ in $Z(C^{\prime})$. So the intersection $Z(C)\cap
Z(C^{\prime})$ is $\\{\pm 1\\}$ by Lemma 138. Thus $C\cap
C^{\prime}\subseteq\\{\pm 1\\}$. ∎

Next we wish to count the number of cyclic subgroups of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ of each order. We have already counted
such groups when the order is $p-1$ or $2p$, so we start with the order $p+1$
case.

######  Lemma 142.

Let $p$ be an odd prime. The number of cyclic subgroup of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ of order $p+1$ is equal to
$\frac{1}{2}(p-1)p$.

###### Proof.

We consider two approaches. One is to note that each element $g$ of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ outside of $\\{\pm 1\\}$ is contained in
a unique maximal cyclic subgroup of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$,
namely the centralizer $Z(g)$ (see Lemma 140). Also a cyclic subgroup of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ is maximal if and only if its order is
$p-1$ (when $p\neq 3$), $2p$, or $p+1$. So we can partition the elements of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ outside of $\\{\pm 1\\}$: if $c_{n}$ is
the number of cyclic subgroups of order $n$ we have

$|{\mathrm{SL}}_{2}({\mathbb{F}}_{p})|-2=c_{p-1}((p-1)-2)+c_{2p}(2p-2)+c_{p+1}((p+1)-2)$

(this works even for $p=3$ since $(p-1)-2=0$). Based on what we know so far

$(p-1)p(p+1)-2=\frac{1}{2}p(p+1)(p-3)+(p+1)(2p-2)+c_{p+1}(p-1).$

Now we solve for $c_{p+1}$.

This first approach is legitmate, but there is also a more group theoretical /
Galois theoretical approach. Let $C$ be any cyclic group of order $p+1$, and
let $E$ be the field $F[C]$ inside of $M_{2}({\mathbb{F}}_{p})$ generated by
the elements of $C$. Then the normalizer $N(C)$ acts via conjugation not just
on $C$ but on all of $E=F[C]$. In fact we get a homomorphism from $N(C)$ to
the Galois group $G$ of $E$ over $F$, and kernel of this homomorphism is just
$Z(C)$. Since $E$ is a quadratic extension of ${\mathbb{F}}_{p}$, the Galois
group $G$ has order 2 (see Lemma 128). So $Z(C)$ has index $1$ or $2$ in
$N(C)$. Lemma 137 shows that $N(C)$ is not $Z(C)$ so the index is actually 2,
and Proposition 140 show that $Z(C)=C$. Thus $[N(C):C]=2$ and so $N(C)$ has
order $2(p+1)$. By the orbit-stabilizer theorem we have that the number of
conjugates of $C$ is

$\frac{(p-1)p(p+1)}{2(p+1)}=\frac{1}{2}(p-1)p$

which counts the number of cyclic subgroups of order $p+1$ since all such
groups are conjugate (Theorem 141). ∎

Our calculations culminate with a full census of the cyclic subgroups of
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$. Since $\mathrm{SL}_{2}({\mathbb{F}}_{p})$
is Sylow-cyloidal, this gives a full census of Abelian subgroups as well.

######  Proposition 143.

Let $p$ be an odd prime and let $c_{m}$ be the number of cyclic subgroups of
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$ of order $m$. Then

$c_{1}=c_{2}=1.$

If $m>2$ divides $p-1$ then

$c_{m}=\frac{1}{2}p(p+1).$

If $m>2$ divides $2q$ then

$c_{m}=p+1.$

If $m>2$ divides $p+1$ then

$c_{m}=\frac{1}{2}p(p-1).$

Otherwise $c_{m}=0$.

###### Proof.

Since there is a unique element of order 2 (Proposition 125) and order 1, we
have $c_{1}=c_{2}=2$. So from now on we assume $m>2$.

Suppose that $m$ divides $p-1$ (so $p>3$ since $m>2$). Each cyclic subgroup
$C$ of order $p-1$ has $Z(C)$ for the maximum cyclic subgroup containing $C$
(Proposition 140). By Corollary 130, $Z(C)$ has order in the set
$\\{p-1,2p,p+1\\}$, and $m$ divides the order of $Z(C)$ since $C$ is a
subgroup of $Z(C)$. Since $m$ divides $p-1$, it cannot divide $2p$ or $p+1$ as
well (since the GCD of $p-1$ with each is $2$ and $m>2$). Thus $Z(C)$ is a
cyclic group of order $p-1$. Conversely each cyclic group of order $p-1$ has a
unique subgroup of order $m$. Thus there is a one-to-one correspondences
between cyclic subgroups of order $m$ and subgroups of order $p-1$. By Lemma
135 there are $\frac{1}{2}p(p+1)$ cyclic subgroups of order $p-1$. Thus
$c_{m}=\frac{1}{2}p(p+1)$.

The case where $m$ divides $2p$ is similar: there is a one-to-one
correspondence $C\mapsto Z(C)$ between cyclic subgroups of order $m$ and
cyclic subgroups of order $2p$. We can use Lemma 136 to conclude that there
are $c_{m}=p+1$ cyclic subgroups of order $m$.

The case where $m$ divides $p+1$ is also similar: there is a one-to-one
correspondence $C\mapsto Z(C)$ between cyclic subgroups of order $m$ and
cyclic subgroups of order $p+1$. We can use Lemma 142 to conclude that there
are $c_{m}=\frac{1}{2}(p-1)p$ cyclic subgroups of order $m$. ∎

######  Corollary 144.

Suppose that $p$ is an odd prime. The only normal cyclic subgroups of
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$ are $\\{1\\}$ and $\\{\pm 1\\}$.

###### Proof.

Suppose that $C$ is a normal subgroup of order $m$. By Theorem 141 all cyclic
subgroups of order $m$ are conjugate. Since $C$ is normal, this means that
there is only one subgroup of order $C$. According to the previous
proposition, this can only happen if $m=1,2$. The result follows now from the
fact that $-1$ is the only element of order two (Proposition 125). ∎

######  Corollary 145.

Suppose that $p$ is an odd prime. Then the only normal Sylow-cyclic subgroups
of $\mathrm{SL}_{2}({\mathbb{F}}_{p})$ are $\\{1\\}$ and $\\{\pm 1\\}$.

###### Proof.

Suppose that $N$ is a normal Sylow-cyclic subgroup of
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$. By Theorem 59 the commutator subgroup
$N^{\prime}$ of $N$ is cyclic of odd order. Since $N^{\prime}$ is
characteristic in $N$ it must be normal in
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$. By the above corollary, $N^{\prime}$ is
the trivial group. So $N$ is Abelian, hence cyclic since $N$ is Sylow-cyclic.
So by the above corollary $N$ is $\\{1\\}$ or $\\{\pm 1\\}$. ∎

######  Lemma 146.

Suppose that $p$ is an odd prime and that $N$ is a normal subgroup of
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$ not equal to $\\{1\\}$ or $\\{\pm 1\\}$.
Then $N$ has odd index in $\mathrm{SL}_{2}({\mathbb{F}}_{p})$.

###### Proof.

Let $N_{2}$ be a $2$-Sylow subgroup of $N$. By the above corollary, $N_{2}$
cannot be cyclic, so $N_{2}$ must be a general quaternion group. In
particular, $N$ contains a cyclic subgroup of order $4$. All cyclic subgroups
of order $4$ are conjugate in $\mathrm{SL}_{2}({\mathbb{F}}_{p})$ by Theorem
141. Since $N$ is normal, it must contain all cyclic subgroups of order $4$.
Let $P_{2}$ be a $2$-Sylow subgroup of $\mathrm{SL}_{2}({\mathbb{F}}_{p})$.
Note that $P_{2}$ is generated by its cyclic subgroups of order $4$. So
$P_{2}$ must be contained in $N$. The result follows. ∎

######  Lemma 147.

Let $p$ be an odd prime. Suppose $\mathrm{SL}_{2}({\mathbb{F}}_{p})$ has a
normal subgroup not equal to $\\{1\\}$ or $\\{\pm 1\\}$ or
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$. Then $\mathrm{SL}_{2}({\mathbb{F}}_{p})$
has a normal subgroup of odd prime index.

###### Proof.

Let $N$ be a normal subgroup not equal to $\\{1\\}$ or $\\{\pm 1\\}$ or
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$. Then $N$ has odd index in
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$ by the previous lemma. Observe that
$G=\mathrm{SL}_{2}({\mathbb{F}}_{p})/N$ is a Sylow-cyclic group of odd order.
By Corollary 56, $G$ has a nontrivial cyclic quotient. Hence $G$ has a
quotient of odd prime order. This means that
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$ has a quotient of odd prime order as well,
and the result follows. ∎

Now we are ready for the third main result:

######  Theorem 148.

Let $p\geq 5$ be an odd prime. Then the normal subgroups of
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$ are $\\{1\\}$, $\\{\pm 1\\}$, and
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$ itself.

###### Proof.

Suppose there are normal subgroups that differ from $\\{1\\}$, $\\{\pm 1\\}$,
and $\mathrm{SL}_{2}({\mathbb{F}}_{p})$ itself. By the previous lemma, there
is a normal subgroup $N$ such that $N$ has prime index $q$ in
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$. Since the order of
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$ is $(p-1)p(p+1)$ we have that $q$ divides
an element of the set $\\{p-1,p,p+1\\}$. It cannot divide two elements of this
set since $q\geq 5$. So there are three cases based on which element is
divisible by $q$.

The next step is to partition $\mathrm{SL}_{2}({\mathbb{F}}_{p})$ as follows:

  * •

Let $\Gamma_{0}=\\{\pm 1\\}$.

  * •

Let $\Gamma_{1}$ be all elements $g$ of order $m\geq 3$ such that the
centralizer $Z(\left<g\right>)$ of $\left<g\right>$ has order $p-1$.

  * •

Let $\Gamma_{2}$ be all elements $g$ of order $m\geq 3$ such that the
centralizer $Z(\left<g\right>)$ of $\left<g\right>$ has order $2p$.

  * •

Let $\Gamma_{3}$ be all elements $g$ of order $m\geq 3$ such that the
centralizer $Z(\left<g\right>)$ of $\left<g\right>$ has order $p+1$.

These sets are clearly disjoint. They form partition of
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$ since every element outside $\Gamma_{0}$
has order at least $3$ (Proposition 125), and $Z(\left<g\right>)$ has order in
the set $\\{p-1,2p,p+1\\}$ (see Proposition 140 and Corollary 130).

Next we further partition each $\Gamma_{i}$. We start with $\Gamma_{1}$. Every
element $g\in\Gamma_{1}$ is in exactly one cyclic subgroup of order $p-1$,
namely $Z(\left<g\right>)$ (it cannot be in a second cyclic group of order
$p-1$ by Theorem 141). So $\Gamma_{1}$ can be partitioned into sets of the
form $C\cap\Gamma_{1}$ where $C$ is a cyclic subgroup of order $p-1$. What is
the size of $C\cap\Gamma_{1}$? By Corollary 130, if $C$ is a cyclic subgroup
of order $p-1$ then it is a maximal cyclic group, and so $C$ equal to
$Z(\left<g\right>)$ for all $g\in C$ of order $m\geq 3$ (Proposition 140). The
only elements of $C$ not in $C\cap\Gamma_{1}$ are the elements of order $1$ or
$2$, namely the elements $\pm 1$. So each partition $C\cap\Gamma_{1}$ has size
$(p-1)-2=p-3$. From Proposition 143 we know that the number of partitions is
$c_{p-1}=\frac{1}{2}p(p+1)$. So $\Gamma_{1}$ has
$c_{p-1}(p-3)=\frac{1}{2}p(p+1)(p-3)$ elements. A similar calculation can be
made for $\Gamma_{2}$ and $\Gamma_{3}$.

In fact, we can list the proportion of elements of
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$ in $\Gamma_{1},\Gamma_{2},\Gamma_{3}$ as
follows:

  * •

The proportion of elements of $\mathrm{SL}_{2}({\mathbb{F}}_{p})$ in
$\Gamma_{1}$ is

$\frac{\frac{1}{2}p(p+1)(p-1-2)}{(p-1)p(p+1)}=\frac{1}{2}\frac{p-3}{p-1}<\frac{1}{2}.$

  * •

The proportion of elements of $\mathrm{SL}_{2}({\mathbb{F}}_{p})$ in
$\Gamma_{2}$ is

$\frac{(p+1)(2p-2)}{(p-1)p(p+1)}=\frac{2}{p}.$

  * •

The proportion of elements of $\mathrm{SL}_{2}({\mathbb{F}}_{p})$ in
$\Gamma_{3}$ is

$\frac{\frac{1}{2}(p-1)p(p+1-2)}{(p-1)p(p+1)}=\frac{1}{2}\frac{p-1}{p+1}<\frac{1}{2}.$

Claim: _the complement of $N$ is contained in $\Gamma_{i}$ for some
$i\in\\{1,2,3\\}$._ For instance, suppose $q$ divides $p-1$. Then clearly
$\Gamma_{0}\subseteq N$. Note that $q$ cannot divide the order of any
$g\in\Gamma_{2}$ since $q$ does not divide $2p$. So each $g\in\Gamma_{2}$ must
have trivial image under
$\mathrm{SL}_{2}({\mathbb{F}}_{p})\to\mathrm{SL}_{2}({\mathbb{F}}_{p})/N$. In
other words, $\Gamma_{2}\subseteq N$. Similarly $q$ cannot divide the order of
any $g\in\Gamma_{3}$ since $q$ does not divide $p+1$. So each $g\in\Gamma_{3}$
must have trivial image under
$\mathrm{SL}_{2}({\mathbb{F}}_{p})\to\mathrm{SL}_{2}({\mathbb{F}}_{p})/N$. In
other words, $\Gamma_{3}\subseteq N$. So the complement of $N$ in
$\mathrm{SL}_{2}({\mathbb{F}}_{p})$ must be contained in $\Gamma_{1}$. The
other cases are similar. In fact, if $N^{c}$ is the complement
$\mathrm{SL}_{2}({\mathbb{F}}_{p})-N$ then

  * •

If $q$ divides $p-1$ then $N^{c}\subseteq\Gamma_{1}$.

  * •

If $q$ divides $p$ then $N^{c}\subseteq\Gamma_{2}$.

  * •

If $q$ divides $p+1$ then $N^{c}\subseteq\Gamma_{3}$.

This leads to a contradiction when $p\geq 5$. Since $N$ is a proper subgroup
of $\mathrm{SL}_{2}({\mathbb{F}}_{p})$, the complement $N^{c}$ must contain at
least one-half of the elements of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$. But
the formulas for proportion of the elements of
$\Gamma_{1},\Gamma_{2},\Gamma_{3}$ in ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$
shows that these proportions are each strictly smaller than one-half. So no
such normal subgroup $N$ of index $q$ exists. ∎

######  Corollary 149.

Let $p\geq 5$ be an odd prime. Then ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ is a
non-solvable Sylow-cycloidal group, and the quotient

$\mathrm{PSL}_{2}({\mathbb{F}}_{p})\;\stackrel{{\scriptstyle\mathrm{def}}}{{\,=\,}}\;{\mathrm{SL}}_{2}({\mathbb{F}}_{p})/\\{\pm
1\\}$

is a simple group.

######  Remark.

The simple groups $\mathrm{PSL}_{2}({\mathbb{F}}_{p})$ were studied even by
Galois, and were the earliest known non-Abelian simple groups outside of the
alternating groups.

The above takes care of the case $p\geq 5$. The case $p=3$ is of interest as
well:

######  Proposition 150.

The group ${\mathrm{SL}}_{2}({\mathbb{F}}_{3})$ is a solvable Sylow-cycloidal
group isomorphic to the binary tetrahedral group $2T$.

###### Proof.

The order of ${\mathrm{SL}}_{2}({\mathbb{F}}_{3})$ is $2\cdot 3\cdot 4=24$
(Proposition 125). By Theorem 131, ${\mathrm{SL}}_{2}({\mathbb{F}}_{3})$ is a
Sylow-cycloidal group whose $2$-Sylow subgroups are not cyclic. Thus the
$2$-Sylow subgroups of ${\mathrm{SL}}_{2}({\mathbb{F}}_{3})$ must be
isomorphic to the quaternion group $Q_{8}$ with $8$ elements.

According to Proposition 143, there are $3$ cyclic subgroups of
${\mathrm{SL}}_{2}({\mathbb{F}}_{3})$ of order $4$ which is exactly the number
of such cyclic subgroups of $Q_{8}$. Thus there is a unique $2$-Sylow subgroup
of ${\mathrm{SL}}_{2}({\mathbb{F}}_{3})$, and so it is normal. This explains
why ${\mathrm{SL}}_{2}({\mathbb{F}}_{3})$ is solvable. Proposition 143 also
asserts that there are $4$ cyclic subgroups of
${\mathrm{SL}}_{2}({\mathbb{F}}_{3})$ of order $3$, and these subgroups are
conjugate (Theorem 141). Thus ${\mathrm{SL}}_{2}({\mathbb{F}}_{3})$ has no
normal subgroup of order 3. By Proposition 102,
${\mathrm{SL}}_{2}({\mathbb{F}}_{3})$ is isomorphic to the binary tetrahedral
group $2T$. ∎

We have considered cyclic subgroups and normal subgroups of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$. Now we will consider subgroups of odd
order. We will do so using the following result on normalizers of cyclic
subgroups.

######  Lemma 151.

Let $C$ be a cyclic subgroup of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ where
$p$ is an odd prime. Then the normalizer $N(C)$ of $C$ is the normalizer
$N(Z(C))$ of the maximal cyclic subgroup $Z(C)$ containing $C$. Furthermore,

  * •

If $Z(C)$ has order $p-1$ then $N(Z(C))$ is a non-Abelian subgroup of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ of order $2(p-1)$.

  * •

If $Z(C)$ has order $2p$ then $N(Z(C))$ is a non-Abelian subgroup of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ of order $(p-1)p$.

  * •

If $Z(C)$ has order $p+1$ then $N(Z(C))$ is a non-Abelian subgroup of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ of order $2(p+1))$.

###### Proof.

Since $Z(C)$ is cyclic, $C$ is a characteristic subgroup of $Z(C)$. Thus $C$
is normal in $N(Z(C))$ since $Z(C)$ is normal in $N(Z(C))$. Hence

$N(Z(C))\subseteq N(C).$

We show equality by showing that both subgroups in this inclusion have the
same order.

We focus on the case where $Z(C)$ has order $p-1$. The other cases are
similar. By Theorem 141 the group ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ acts
transitively on the collection of cyclic subgroups of order $|C|$. By
Proposition 143 the orbit of $C$ under this action has size
$\frac{1}{2}p(p+1)$. So by the orbit-stabilizer theorem, the stabilizer $N(C)$
under conjugation has the following size:

$|N(C)|=\frac{(p-1)p(p+1)}{\frac{1}{2}p(p+1)}=2(p-1).$

This calculation is valid for $N(Z(C))$ as well since $Z(C)$ is cyclic of size
$p+1$. So $N(Z(C))$ has order $2(p-1)$ as well. Equality follows. ∎

######  Corollary 152.

Suppose $H$ is a noncyclic subgroup of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$
of odd order, where $p$ is an odd prime. Then $H$ has order dividing $(p-1)p$.
Furthermore $H$ contains a normal subgroup $C$ of order $p$, and $H$ is
subgroup of the normalizer $N(C)$ of $C$ in
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$.

###### Proof.

Since ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ is a Sylow-cycloidal group, the
subgroup $H$ of odd order is a Sylow-cyclic group. By Theorem 59, $H$ has a
nontrivial normal cyclic subgroup $C$, and so $H\subseteq N(C)$. If $C$ has
order dividing $p-1$ or $p-2$ then $N(C)/Z(C)$ is even, and the image of $H$
in $N(C)/Z(C)$ is trivial. Thus $H\subseteq Z(C)$ and so $H$ is cyclic. So we
are left with the case where $C$ has order dividing $2p$. In other words, $C$
has order $p$. By the above proposition $N(C)$ has order $(p-1)p$. ∎

Given the above corollary, the normalizer $N(C)$ of a cyclic subgroup $C$ of
order $p$ warrants our further attention.

######  Lemma 153.

Let $C_{p}$ be a subgroup of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ of order
$p$, where $p$ is an odd prime. Then $C_{p}$ is conjugate to the subgroup

$\left\\{\begin{pmatrix}1&b\\\
0&1\end{pmatrix}\;\middle|\;b\in{\mathbb{F}}_{p}\right\\}\cong{\mathbb{F}}_{p}.$

The normalizer $N(C_{p})$ is conjugate to the subgroup

$\left\\{\begin{pmatrix}a&b\\\
0&a^{-1}\end{pmatrix}\;\middle|\;a\in{\mathbb{F}}^{\times}_{p},\;b\in{\mathbb{F}}_{p}\right\\}.$

Furthermore, $N(C_{p})$ is a semidirect product $C_{p}\rtimes C_{p-1}$ where
$C_{p-1}$ is the cyclic group

$\left\\{\begin{pmatrix}a&0\\\
0&a^{-1}\end{pmatrix}\;\middle|\;a\in{\mathbb{F}}^{\times}_{p}\right\\}\cong{\mathbb{F}}_{p}^{\times}.$

Viewing $C_{p-1}$ as ${\mathbb{F}}_{p}^{\times}$ and $C_{p}$ as
${\mathbb{F}}_{p}$, the action of $a\in C_{p-1}$ on $C_{p}$ is given by

$b\mapsto a^{2}b.$

###### Proof.

Let $\alpha$ be a generator of $C_{p}$. By Lemma 129, $\alpha$ has exactly one
eigenvalue in ${\mathbb{F}}_{p}$. If we work in a basis $v_{1},v_{2}$ where
$v_{1}$ is an eigenvector of $\alpha$, then $\alpha$ has the form

$\begin{pmatrix}1&b\\\ 0&1\end{pmatrix}$

for some nonzero $b\in{\mathbb{F}}_{p}$. It follows that $C_{p}$ has the
desired form (up to conjugation). From now on we assume $C_{p}$ is this
particular subgroup of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$.

Next consider the subgroup

$H\;\stackrel{{\scriptstyle\mathrm{def}}}{{\,=\,}}\;\left\\{\begin{pmatrix}a&b\\\
0&a^{-1}\end{pmatrix}\;\middle|\;a\in{\mathbb{F}}^{\times}_{p},\;b\in{\mathbb{F}}_{p}\right\\}.$

Observe that

$\begin{pmatrix}a&b\\\ 0&a^{-1}\end{pmatrix}\mapsto a$

is a surjective homomorphism $H\to{\mathbb{F}}_{p}^{\times}$ with kernel
$C_{p}$. So $C_{p}$ is a normal subgroup of $H$, and so $H\subseteq N(C_{p})$.
By the above lemma $N(C_{p})$ has size $(p-1)p$, which is also the size of
$H$. So $H=N(C_{p})$. Note that $C_{p-1}$ (as defined in the statement of the
lemma) is a complement to $C_{p}$ in $H=N(C_{p})$ and so $N(C_{p})$ is a
semidirect product $C_{p}\rtimes C_{p-1}$. The last assertion follows from the
calculation:

$\begin{pmatrix}a&0\\\ 0&a^{-1}\end{pmatrix}\begin{pmatrix}1&b\\\
0&1\end{pmatrix}\begin{pmatrix}a&0\\\
0&a^{-1}\end{pmatrix}^{-1}=\begin{pmatrix}a&0\\\
0&a^{-1}\end{pmatrix}\begin{pmatrix}1&b\\\
0&1\end{pmatrix}\begin{pmatrix}a^{-1}&0\\\
0&a\end{pmatrix}=\begin{pmatrix}1&b\\\ 0&1\end{pmatrix}$

∎

Here is the next major result:

######  Theorem 154.

Let $H$ be a noncyclic subgroup of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ of
odd order, where $p$ is an odd prime. Then $H$ contains a cyclic subgroup of
order $p$ and $H$ is contained in the normalizer $N(C)$ of $C$ in
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$. Each such $H$ is isomorphic to
${\mathbb{F}}_{p}\rtimes T$ where $T$ is a nontrivial subgroup of
${\mathbb{F}}_{p}^{\times}$ of odd order. Here the action of $a\in T$ is given
by $b\mapsto a^{2}b$.

Furthermore, given such a semidirect product, ${\mathbb{F}}_{p}\rtimes T$
there is a non-Abelian subgroup of ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$
isomorphic to ${\mathbb{F}}_{p}\rtimes T$.

###### Proof.

The existence of $C$ follows from Corollary 152. By the above lemma, any such
$H$ is isomorphic to a subgroup of
${\mathbb{F}}_{p}\rtimes{\mathbb{F}}_{p}^{\times}$ where ${\mathbb{F}}_{p}$
corresponds to $C$. Observe that $H$ is actually isomorphism to
${\mathbb{F}}_{p}\rtimes T$ where $T$ is the image of $H$ in
${\mathbb{F}}_{p}^{\times}$. Note that $T$ is nontrivial since $H$ is
noncyclic.

Conversely, given a nontrivial odd order subgroup $T$ of
${\mathbb{F}}_{p}^{\times}$, we have a subgroup of $N(C)$ isomorphic
${\mathbb{F}}_{p}\rtimes T$ where here $C$ is any subgroup of
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ of order $p$. (This follows from the
above lemma). ∎

This leads to some important corollaries:

######  Corollary 155.

Suppose $p$ is an odd prime. Then ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$
contains a noncyclic subgroup of order the product of two primes if and only
if $p$ is not a Fermat prime.

###### Proof.

First suppose ${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ has a noncyclic subgroup
$H$ of order $|H|=q_{1}q_{2}$ where $q_{1}\leq q_{2}$ are primes. Since
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ is a Sylow-cycloidal group, any subgroup
of prime squared order is cyclic. Thus $q_{1}\neq q_{2}$.

First suppose that $q_{1}=2$. By Theorem 59 we have that $H$ is isomorphic to
a semidirect product $C_{q}\rtimes C_{2}$ where $C_{q}$ and $C_{2}$ are cyclic
subgroups of order $q=q_{2}$ and $2$ respectively. The only nontrivial action
of $C_{2}$ is the one where the nontrivial element of $C_{2}$ acts on $C_{q}$
by $x\mapsto x^{-1}$. Thus $H$ is a dihedral group of order $2q$, and so
contains $q$ elements of order 2. However,
${\mathrm{SL}}_{2}({\mathbb{F}}_{p})$ has a unique element of order $2$. Thus