# An Upper Bound of Composition Series111Email<EMAIL_ADDRESS>
Abhijit Bhattacharjee

( Department of Mathematics, Institute of Science, Banaras Hindu University,
Varanasi, India-221005.
August 27, 2024)

###### Abstract

In this paper we prove that among all finite groups of order $\leq n$ (where
$n\geq 4$ be a natural number) the number of distinct composition series is
bounded above by $\prod_{i=1}^{\left[\log_{2}n\right]}\left(2^{i}-1\right)$
and it is attained if and only if $G$ is the elementary abelian $2$-group of
order $2^{\alpha},$ where $\alpha=\left[\log_{2}n\right]$. This bound is the
first ever non trivial upper bound of composition series and so far best
possible.

Keywords: Finite group, composition series, Sylow subgroup, elementary abelian
group

Mathematics Subject Classifications(2010): 20D30, 20D15

##  1  Introduction

A composition series, that is a series of subgroups each normal in the
previous such that corresponding factor groups are simple. Any finite group
has a composition series.The concept of a composition series in a group is due
to Evariste Galois(1831)[Theorem 5.9, [1]]. The famous Jordan-Holder theorem
which says that, the composition factors in a composition series are unique up
to isomorphism was proved in nineteenth century [2, 3]. Sometimes a group of
small order has a huge number of distinct composition series. For example an
elementary abelian group of order 64 has 615195 distinct composition series.To
find an upper bound for composition series in a finite group is a natural
question. In [4], there is an algorithm in GAP to find the distinct
composition series of any group of finite order.The aim of this paper is to
provide an upper bound of the number of distinct composition series of any
group of finite order. The approach of this is combinatorial and the method is
elementary. All the groups considered in this paper are of finite order. There
are three sections in the paper namely section 2, section 3 and section 4. In
section 2 we prove Theorem 2A which is the main theorem of section 2. All the
theorems and results in section 2 are well known. In section 3 we prove
Theorem 3B which is a new result and it is the main theorem of section 3. In
section 4 we prove Theorem 4C by proving a new inequality in number theory.
This theorem is the main theorem of this paper.

##  2 Preliminary Results

######  Theorem 2.1.

Among all finite groups of order $n$, the abelian group with elementary
abelian Sylow subgroups, has the highest number of distinct composition
series.

Before proving theorem 2.1 we recall some useful theorems in group theory at
first.

######  Theorem 2.2 (Theorem 2.28, [1]).

(Correspondence Theorem) Given $G\triangleright N$, denote the the family of
subgroups of $G$ that contain $N$ by $sub(G;N)$ and the family of subgroups of
$G/N$ by $sub(G/N)$. Then there is a bijection

$\phi:sub(G;N)\longrightarrow sub(G/N),$ $H\mapsto H/N$

which preserves all subgroup lattice and normality relationships.

######  Theorem 2.3 (Exercise 14, Section 5.2, [5]).

A finite abelian group is self dual.

######  Lemma 2.4 (Lemma 4.2.7, [6]).

Suppose $G$ be a finite group and let
$\prod\left(G\right)=\left\\{p_{1},p_{2},\dots,p_{r}\right\\}$ where
$\prod\left(G\right)$ denotes the distinct prime factors in the order of $G$.
Let $P_{i}\in Syl_{p_{i}}\left(G\right)$. If $G=P_{1}\times
P_{2}\times\dots\times P_{r}$ then

$\mid m\left(G\right)\mid=\sum_{i=1}^{r}\mid m\left(P_{i}\right)\mid,$

where $m\left(G\right)$denotes the distinct maximal subgroups in $G$.

######  Corollary 2.5.

If $G$ be the abelian group with elementary abelian Sylow subgroups of order
$n$ where $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\dots p_{r}^{\alpha_{r}}$
then
$m\left(G\right)=\sum_{i=1}^{r}\left(\frac{p_{i}^{\alpha_{i}}-1}{p_{i}-1}\right)$.

###### Proof.

Since $G$ is a finite abelian group then it is self dual and direct products
of its Sylow subgroups. Therefore number of maximal subgroups in $G$ is equal
to the sum of all the maximal subgroups in it’s Sylow subgroups. Now
elementary abelian group is self dual. So number of maximal subgroups in an
elementary abelian group is exactly same with the number of minimal
subgroups.Therefore number of subgroups of index $p_{i}$ is number of distinct
subgroups of order $p_{i}$in an elementary abelian $p_{i}$group of order
$p_{i}^{\alpha_{i}}$ which is $\frac{p_{i}^{\alpha_{i}}-1}{p_{i}-1}$ and the
result follows immediately. ∎

######  Example 2.6.

If
$G=\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{3}\times\mathbb{Z}_{3}\times\mathbb{Z}_{5}\times\mathbb{Z}_{5}$
then $\mid G\mid=3600$ and

$m\left(G\right)=\frac{2^{4}-1}{2-1}+\frac{3^{2}-1}{3-1}+\frac{5^{2}-1}{5-1}=15+4+6=25.$

######  Lemma 2.7.

If $G=S_{1}\times S_{2}$, where $S_{1}$, $S_{2}$ are two finite simple groups.
If $G_{1}$ is a proper non-trivial normal subgroup of $G$ then $G_{1}$ is
isomorphic to either $S_{1}$or $S_{2}$.

###### Proof.

We consider the projection map $p:G\rightarrow S_{1}$ and the projection map
is onto. Therefore the image of a normal subgroup under the projection map $p$
will be a normal subgroup in $S_{1}$. Since $S_{1}$ is a simple group we have
the following two choices : Case 1 : If
$p\left(G_{1}\right)=\left\\{e\right\\}$. Then $G_{1}$ is a subgroup of
$S_{2}$ and it is a non trivial normal subgroup of the simple group group
$S_{2}$ then $G_{1}=S_{2}$ i.e. $G_{1}{\backsimeq}S_{2}$. Case 2 : If
$p\left(G_{1}\right)=S_{1}$. Now we will take the kernel $K=S_{2}\cap G_{1}$of
the projection map. Here $K$ will become a normal subgroup of the simple group
$S_{2}$. So either $K=\left\\{e\right\\}$or $S_{2}$. If $K=\left\\{e\right\\}$
then $G_{1}\backsimeq S_{1}$. If $K=S_{2}$ then $G_{1}\backsimeq S_{1}\times
S_{2}$, which is a contradiction since $G_{1}$is a proper subgroup of $G$. ∎

######  Corollary 2.8.

If $G=S_{1}\times S_{2}\times\dots\times S_{k}$ be the direct product of $k$
non-abelian simple groups then $G$ has $2^{k}$ normal subgroups namely
$N_{1}\times N_{2}\times\dots\times N_{k}$ where either
$N_{i}=\left\\{e\right\\}$ or $N_{i}=S_{i}$ for $1\leq i\leq k$. Also $G$ has
only $k$ maximal normal subgroups namely $M_{1},M_{2},\dots,M_{k}$ where
$M_{i}=S_{1}\times S_{2}\times\dots\times S_{i-1}\times e\times
S_{i+1}\times\dots\times S_{k}$ and $G/M_{i}\backsimeq S_{i}$ for $1\leq i\leq
k$.

######  Example 2.9.

Let $G=A_{5}\times A_{5}$ then $\mid G\mid=3600$ and $G$ has only $2^{2}=4$
normal subgroups namely $e\times e$, $e\times A_{5}$, $A_{5}\times e$, $G$ and
it has only $2$ maximal normal subgroups namely $e\times A_{5}$ and
$A_{5}\times e$.

Now we present the proof of theorem 2.1.

###### Proof.

Let $G$ be the abelian group with elementary abelian Sylow subgroups such that
$\mid G\mid=n$. Now by lemma 2.4, number of maximal subgroups in a finite nil-
potent group of order $n$ is exactly same of the sum of distinct maximal
subgroups in each Sylow subgroup. But we know that among all finite $p$ groups
of same order the elementary abelian group has highest number of maximal
subgroups. This proves that among all nil-potent groups of order $n$, $G$ has
the highest number of maximal normal subgroups and hence the highest number of
composition series. Let $K$ be a solvable group of order $n$ with $m$ maximal
normal subgroups namely $M_{1},M_{2},\dots,M_{m}$. Then index of $M_{i}$ in
$K$ is a prime and $K/M_{i}\backsimeq\mathbb{Z}_{p}$ for some prime $p$. Let
$C^{\prime}$ be the commutator subgroup of $K$. Then
$C^{\prime}\trianglelefteq M_{i}.$ Let $M=\cap_{i=1}^{m}M_{i}$ be the
intersection of all the maximal normal subgroups of $K$. $M$ is a normal
subgroup of $K$ such that $C^{\prime}\trianglelefteq M$ and therefore $K/M$ is
abelian. Now by Correspondence Theorem there exists a bijection

$\phi:sub\left(K;M\right)\longrightarrow sub\left(K/M\right),$

which preserves subgroup lattice and normality relationship. Therefore $K/M$
is an abelian group with exactly $m$ maximal normal subgroups. But it is
already proved that among all abelian groups of same order the abelian group
with elementary abelian Sylow subgroups, has the highest number of maximal
normal subgroups.For non solvable non simple group we take the Jacobson
radical $J(G)$ which is the intersection of all the maximal normal subgroups
of $G$ . Then $J(G)$ is a characteristic subgroup of $G$ and the smallest
normal subgroup of $G$ such the the quotient $G/J(G)$ is the direct product of
simple groups. Now if $J(G)$ is trivial then $G$ is direct product of simple
groups. Otherwise, let $J(G)$ be a non trivial then we can get a minimal
normal subgroup $P$ such that $P$ is a subgroup of $J(G)$ . Then we take the
quotient $G/P$ and get a normal covering. Proceeding this way it reduces to
the case when $G$ is a direct products of simple groups as $G$ embeds in the
direct product of quotient of $k$ maximal normal subgroups such that the
intersection of these $k$ maximal normal subgroups becomes trivial, where $k$
is a natural number. Now if $H$ is a non-solvable non-simple group of order
$n$ then $H$ is direct product of isomorphic non-abelian simple groups by
Corollary 2.7, direct product of $k$ copies of non-abelian simple groups has
only $k$ maximal normal subgroups, which proves the theorem. ∎

##  3 Some New Results

In this section we prove Theorem 3.7, which is the main result in this
section.

######  Definition 3.1.

We define $C_{G}$ as the set of all distinct composition series of the group
$G$.

######  Theorem 3.2.

Let $n\geq 2$ be a positive integer where
$n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\dots p_{r}^{\alpha_{r}}$. Then

$\mid
C_{\mathbb{Z}_{n}}\mid=\frac{\left(\sum_{i=1}^{r}\alpha_{i}\right)!}{\prod_{i=1}^{r}\alpha_{i}!}.$

###### Proof.

We know that for each positive divisor $d$ of $n$ there exists a unique
subgroup of $\mathbb{Z}_{n}$ of order $n$. Let $S$ be the set of sequence of
primes from the set $\\{p_{1},\dots,p_{r}\\}$ of length
$\alpha_{1}+\alpha_{2}+\dots+\alpha_{r}$, where $p_{i}$ occurs $\alpha_{i}$
times. Then

$\mid
S\mid=\frac{\left(\sum_{i=1}^{r}\alpha_{i}\right)!}{\prod_{i=1}^{r}\alpha_{i}!}.$

Now $C_{\mathbb{Z}_{n}}$ is the set of distinct composition series of
$\mathbb{Z}_{n}$. Now define a function $f:S\rightarrow C_{\mathbb{Z}_{n}}$ by

$f\left(\beta_{1}\beta_{2}\dots\beta_{\alpha_{1}+\alpha_{2}+\dots+\alpha_{r-1}}\beta_{\alpha_{1}+\alpha_{2}+\dots+\alpha_{r}}\right)=\left\\{e\right\\}\trianglelefteq\mathbb{Z}_{\beta_{1}}\trianglelefteq\mathbb{Z}_{\beta_{1}\beta_{2}}\trianglelefteq\dots\trianglelefteq\mathbb{Z}_{\beta_{1}\beta_{2}\dots\beta_{\alpha_{1}+\alpha_{2}+\dots+\alpha_{r}}}=\mathbb{Z}_{n},$

where
$\beta_{1},\beta_{2},\dots,\beta_{\alpha_{1}+\alpha_{2}+\dots+\alpha_{r}}$ are
primes such that $\beta_{i}=p_{i}$ has exactly $\alpha_{i}$ solutions for
$1\leq i\leq\alpha_{1}+\alpha_{2}+\dots+\alpha_{r}$. So
$\beta_{1}\beta_{2}\dots\beta_{\alpha_{1}+\alpha_{2}+\dots+\alpha_{r-1}}\beta_{\alpha_{1}+\alpha_{2}+\dots+\alpha_{r}}\in
S$. Now
$\left\\{e\right\\}\trianglelefteq\mathbb{Z}_{\beta_{1}}\trianglelefteq\mathbb{Z}_{\beta_{1}\beta_{2}}\trianglelefteq\dots\trianglelefteq\mathbb{Z}_{\beta_{1}\beta_{2}\dots\beta_{\alpha_{1}+\alpha_{2}+\dots+\alpha_{r}}}=\mathbb{Z}_{n}$
i.e.$\left\\{e\right\\}\trianglelefteq\mathbb{Z}_{\theta_{1}}\trianglelefteq\mathbb{Z}_{\theta_{2}}\trianglelefteq\dots\trianglelefteq\mathbb{Z}_{\theta_{\alpha_{1}+\alpha_{2}+\dots+\alpha_{r}}}=\mathbb{Z}_{n}$
is a composition series of $\mathbb{Z}_{n}$ where
$\theta_{i}=\prod_{j=1}^{i}\beta_{j}$, $1\leq
i\leq\alpha_{1}+\alpha_{2}+\dots+\alpha_{r}.$ Then $f:S\rightarrow
C_{\mathbb{Z}_{n}}$ is an injective mapping by its construction. Now we will
prove that it is surjective also. Let $\left\\{e\right\\}=G_{0}\trianglelefteq
G_{1}\trianglelefteq\dots\trianglelefteq
G_{\alpha_{1}+\alpha_{2}+\dots+\alpha_{r}}=\mathbb{Z}_{n}$ be a composition
series of $\mathbb{Z}_{n}.$ Since for each divisor $d$ of $n$, there exists a
unique subgroup of order $d$ of $\mathbb{Z}_{n}$ and any subgroup of
$\mathbb{Z}_{n}$ is cyclic then $\dfrac{\mid G_{i}\mid}{\mid G_{i-1}\mid}$ is
a prime number. Define $g:C_{\mathbb{Z}_{n}}\longrightarrow S$ by

$g\left(C_{\mathbb{Z}_{n}}\right)=q_{1}q_{2}\dots
q_{\alpha_{1}+\alpha_{2}+\dots+\alpha_{r}},$

where $q_{i}=\dfrac{\mid G_{i}\mid}{\mid G_{i-1}\mid},$ for $1\leq
i\leq\alpha_{1}+\alpha_{2}+\dots+\alpha_{r}.$ Then each $q_{i}$ is a prime for
$1\leq i\leq\alpha+\alpha_{2}+\dots+\alpha_{r}.$ So $q_{1}q_{2}\dots
q_{\alpha_{1}+\alpha_{2}+\dots+\alpha_{r}}$ is a sequence of primes such that
$q_{i}=p_{i}$ has exactly $\alpha_{i}$ solutions for $1\leq
i\leq\alpha_{1}+\alpha_{2}+\dots+\alpha_{r}$. So, $q_{1}q_{2}\dots
q_{\alpha_{1}+\alpha_{2}+\dots+\alpha_{r}}\in S$ and therefore $f^{-1}=g$ and
hence $f$ is surjective and hence bijective also. Therefore,

$\mid S\mid=\mid
C_{\mathbb{Z}_{n}}\mid=\frac{\left(\sum_{i=1}^{r}\alpha_{i}\right)!}{\prod_{i=1}^{r}\alpha_{i}!}$

i.e. $\mathbb{Z}_{n}$ has
$\dfrac{\left(\sum_{i=1}^{r}\alpha_{i}\right)!}{\prod_{i=1}^{r}\alpha_{i}!}$
distinct composition series. ∎

######  Example 3.3.

Let us try to understand the above theorem for $\mathbb{Z}_{360}$. Note that,
$360=2^{3}3^{2}5^{1}$. Now $\frac{\left(3+2+1\right)!}{3!2!1!}=60$ distinct
sequence of primes can be formed of length $6$ such that each sequence
contains $3$ times prime $2$,$2$ times prime $3$ and $1$ time prime $5$. Some
of them are $232253,323522,523322.$ Now we will construct a composition series
corresponding to these numbers.

  1. 1.

The composition series corresponding to $232253$ will be

$\left\\{e\right\\}\trianglelefteq\mathbb{Z}_{2}\trianglelefteq\mathbb{Z}_{6}\trianglelefteq\mathbb{Z}_{12}\trianglelefteq\mathbb{Z}_{24}\trianglelefteq\mathbb{Z}_{120}\trianglelefteq\mathbb{Z}_{360}.$

  2. 2.

The composition series corresponding to $323522$ will be

$\left\\{e\right\\}\trianglelefteq\mathbb{Z}_{3}\trianglelefteq\mathbb{Z}_{6}\trianglelefteq\mathbb{Z}_{18}\trianglelefteq\mathbb{Z}_{90}\trianglelefteq\mathbb{Z}_{180}\trianglelefteq\mathbb{Z}_{360}.$

  3. 3.

The composition series corresponding to $523322$ will be

$\left\\{e\right\\}\trianglelefteq\mathbb{Z}_{5}\trianglelefteq\mathbb{Z}_{10}\trianglelefteq\mathbb{Z}_{30}\trianglelefteq\mathbb{Z}_{90}\trianglelefteq\mathbb{Z}_{180}\trianglelefteq\mathbb{Z}_{360}.$

######  Theorem 3.4.

Let $G$ be an abelian group of order $n$ where $n\geq 2$ be a natural number
such that $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\dots p_{r}^{\alpha_{r}}.$
Then $\mid
C_{G}\mid=\prod_{i=1}^{r}t_{i}\frac{\left(\sum_{i=1}^{r}\alpha_{i}\right)!}{\prod_{i=1}^{r}\alpha_{i}!}$
where $t_{i}$ are the numbers of distinct composition series of the Sylow
$p_{i}$\- subgroups of $G.$

###### Proof.

Note that in finite abelian group there exists at least one subgroup of order
$h$ for each divisor $h$ of the order of the group $n$ and also each
composition factor of an abelian group is group of prime order. The order of
composition factors of an finite abelian group is a prime number and using
them we can make a sequence of primes which belongs to $S$ ( which is
described in the Theorem 3.2 ).

Therefore the number of distinct composition series of $G$ is a multiple of
$\frac{\left(\sum_{i=1}^{r}\alpha_{i}\right)!}{\prod_{i=1}^{r}\alpha_{i}!}.$
Now let $\left\\{e\right\\}=G_{0}\trianglelefteq
G_{1}\trianglelefteq\dots\trianglelefteq G_{m}\trianglelefteq
G_{m+1}\trianglelefteq\dots\trianglelefteq
G_{\alpha_{1}+\alpha_{2}+\dots+\alpha_{r}}=G$ be a composition series of $G$
where $j$ be the least positive integer such that $p_{1}^{s_{1}}$ divide $\mid
G_{j}\mid$and $k$ be the least positive integer such that $p_{1}^{s_{1}+1}$
divide $\mid G_{k}\mid$ when

$1\leq j<k\leq\alpha_{1}+\alpha_{2}+\dots+\alpha_{r},\leavevmode\nobreak\
1\leq s<s+1\leq\alpha_{1}.$

Therefore $\mid G_{j}\mid=p_{1}^{s_{1}}q$ where $p_{1}^{s_{1}},q\in\mathbb{N}$
with $\gcd\left(p_{1}^{s_{1}},q\right)=1.$ $\mid G_{k}\mid=p_{1}^{s_{1}+1}u$,
where $p_{1}^{s_{1}+1},u\in\mathbb{N}$ with
$\gcd\left(p_{1}^{s_{1}+1},u\right)=1.$ Let $G_{j}=A_{j}B_{j}$ where $A_{j}$
and $B_{j}$ are two subgroups of $G$ of order $p_{1}^{s}$ and $q$
respectively. Let $G_{k}=C_{k}D_{k}$ where $C_{k}$ and $D_{k}$ are two
subgroups of $G$ of order $p_{1}^{s_{1}+1}$ and $u$ respectively. We will
prove that $A_{j}\subseteq C_{k}$ ; if not then, $\exists$ some $x\in A_{j}$
such that $x\notin C_{k}.$ $x\in A_{j}\Rightarrow x=p_{1}^{v_{1}}$. Now $x\in
A_{j}\Longrightarrow x\in G_{j,}$ $x\in G_{j}\Longrightarrow x\in G_{j+1}$ as
$G_{j}$ is a normal subgroup of $G_{j+1}$. $x\in G_{j+1}\Longrightarrow x\in
G_{j+2}$ as $G_{j+1}$ is a normal subgroup of $G_{j+2}$. So proceeding in this
way we can prove that $x\in G_{k}$. Now $G_{k}=C_{k}D_{k}$ where $C_{k}\cap
D_{k}=\left\\{e\right\\}.$ $x\in G_{k}\Longrightarrow x\in C_{k}D_{k}$, i.e.
$x=cd$ where $c\in C_{k},d\in D_{k}$. Therefore
$x=p_{1}^{c_{1}}d,p_{1}^{c_{1}}\in C_{k}$. So $p_{1}^{v_{1}}=p_{1}^{c_{1}}d$
i.e. $p_{1}^{v_{1}-c_{1}}=d,$ As $\gcd\left(p_{1}^{\alpha_{1}},u\right)=1$ so
$P_{1}\cap D_{k}=\left\\{e\right\\}$ but $p_{1}^{v_{1}-c_{1}}\in P_{1}\cap
D_{k}$ i.e. $p_{1}^{v_{1}-c_{1}}=e$,$p_{1}^{v_{1}}=p_{1}^{c_{1}}$ hence
$x\left(=p_{1}^{c_{1}}\right)\in C_{k}$, a contradiction. Therefore in the
composition series of Sylow $p_{1}$-group, $A_{j}$ is a subgroup of $C_{k}$.
So if the Sylow $p_{1}$-group has $t_{1}$ distinct composition series then we
have
$t_{1}\frac{\left(\sum_{i=1}^{r}\alpha_{i}\right)!}{\prod_{i=1}^{r}\alpha_{i}!}$
such possible composition series due to Sylow $p_{1}$-subgroup in $G$. In a
similar way for the other Sylow $p_{2}$-group, …, Sylow $p_{r}$\- group we
have
$\prod_{i=1}^{r}t_{i}\frac{\left(\sum_{i=1}^{r}\alpha_{i}\right)!}{\prod_{i=1}^{r}\alpha_{i}!}$
distinct composition series for $G$. ∎

######  Theorem 3.5.

In an elementary abelian group of order $p^{n}$, there are
$\prod_{k=1}^{n}\left(\frac{p^{k}-1}{p-1}\right)=\frac{\prod_{k=1}^{n}\left(p^{k}-1\right)}{\left(p-1\right)^{n}}$
distinct composition series.

###### Proof.

Let $\left\\{e\right\\}\trianglelefteq p\trianglelefteq
p^{2}\trianglelefteq\dots\trianglelefteq p^{k-1}\trianglelefteq
p^{k}\trianglelefteq\dots\trianglelefteq p^{n}$ be a composition series for
the elementary abelian group of order $p^{n}$. To find the number of distinct
composition series, we will find the possible choice from $p^{k-1}$ to $p^{k}$
at first.

Let there be $C_{k-1}$ choice available for $p^{k-1}$ to $p^{k}$. Now in an
elementary abelian group of order $p^{n}$ there are

$\frac{\prod_{i=0}^{k-1}\left(\left(p^{n}-1\right)-\left(p^{i}-1\right)\right)}{\prod_{i=0}^{k-1}\left(p^{k}-p^{i}\right)}=A\leavevmode\nobreak\
\mbox{(say)}$

distinct subgroups of order $p^{k}$ each of which is an elementary abelian (By
[7], page - 65). In an elementary abelian group of order $p^{n}$, also there
are

$\frac{\prod_{i=0}^{k-2}\left(\left(p^{n}-1\right)-\left(p^{i}-1\right)\right)}{\prod_{i=0}^{k-2}\left(p^{k}-p^{i}\right)}=B\leavevmode\nobreak\
\mbox{(say)}$

distinct subgroups of order $p^{k-1}$ each of which is an elementary
abelian(By [7], page-65). Again in an elementary abelian group of order
$p^{k}$ there are

$\frac{\prod_{i=0}^{k-2}\left(\left(p^{k}-1\right)-\left(p^{i}-1\right)\right)}{\prod_{i=0}^{k-2}\left(p^{k}-p^{i}\right)}=D\leavevmode\nobreak\
\mbox{(say)}$

distinct subgroups of order $p^{k-1}$ each of which is elementary abelian (By
[7], page-65). Now there is an identical algebraic structure of any two
subgroups of same order in an elementary abelian group i.e. in elementary
abelian group of order $p^{n}$ , any two distinct subgroups of order $p^{k}$
have the same number of subgroup order $p^{k-1}$. Then

$C_{k-1}=\frac{A.B}{D}=\frac{p^{n+1-k}-1}{p-1}.$

As a result we have
$\prod_{k=1}^{n-1}\left(\frac{p^{n+1-k}-1}{p-1}\right)$distinct possible
choice for the composition series of an elementary abelian group of order
$p^{n}$ ( as $k\geq 1$ ), which is same as

$\prod_{k=1}^{n}\left(\frac{p^{n+1-k}-1}{p-1}\right)=\prod_{k=1}^{n}\left(\frac{p^{k}-1}{p-1}\right)=\frac{\prod_{k=1}^{n}\left(p^{k}-1\right)}{\left(p-1\right)^{n}}.$

∎

######  Theorem 3.6.

Let $G$ be an abelian group with elementary abelian Sylow subgroups of order
$n\geq 2$ where $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\dots
p_{r}^{\alpha_{r}}$, then

$\mid
C_{G}\mid=\prod_{i=1}^{r}\left(\prod_{j=1}^{\alpha_{i}}\frac{p_{i}^{j}-1}{p_{i}-1}\right)\frac{\left(\sum_{i=1}^{r}\alpha_{i}\right)!}{\prod_{i=1}^{r}\alpha_{i}!}.$

###### Proof.

This followed from theorem 3.4 and theorem 3.5. ∎

######  Theorem 3.7.

Let $G$ be a finite group of order $n$ where
$n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\dots p_{r}^{\alpha_{r}}$, then

$\mid
C_{G}\mid\leq\prod_{i=1}^{r}\left(\prod_{j=1}^{\alpha_{i}}\frac{p_{i}^{j}-1}{p_{i}-1}\right)\frac{\left(\sum_{i=1}^{r}\alpha_{i}\right)!}{\prod_{i=1}^{r}\alpha_{i}!}$

and the equality is hold if and only if $G$ is the abelian group with
elementary abelian sylow subgroups.

###### Proof.

This followed from theorem 2.1 and theorem 3.6. ∎

##  4 Main Result

In this section we prove theorem 4.3 which is the main theorem of the paper.
This section is more elementary and computational.

######  Lemma 4.1.

Let $X=\prod_{i=\alpha_{1}+1}^{\alpha_{1}+k}(2^{i}-1)\alpha_{1}!\alpha_{r}!$
and $Y=\prod_{j=1}^{\alpha_{r}}(\frac{p^{j}-1}{p-1})(\alpha_{1}+\alpha_{r})!$,
where $p\geq 5$ is a prime and $\alpha_{1}\geq 0$ is an integer,
$\alpha_{r}\in\mathbb{N}$, such that $t=p^{\alpha_{r}}$ and $[\log_{2}t]=k$.
Also $\alpha_{r}\geq 2$ whenever $p=3$. Then $\frac{X}{Y}$ is a monotone
increasing function of $\alpha_{1}$.

###### Proof.

Let

$f(\alpha_{1})=\frac{\prod_{i=1}^{\alpha_{1}+k}(2^{i}-1)}{\prod_{i=1}^{\alpha_{1}}(2^{i}-1)\prod_{j=1}^{\alpha_{r}}(\frac{p^{j}-1}{p-1})\frac{(\alpha_{1}+\alpha_{r})!}{\alpha_{1}!\alpha_{r}!}}.$

Then

$f(\alpha_{1}+1)=\frac{\prod_{i=1}^{\alpha_{1}+k+1}(2^{i}-1)}{\prod_{i=1}^{\alpha_{1}+1}(2^{i}-1)\prod_{j=1}^{\alpha_{r}}(\frac{p^{j}-1}{p-1})\frac{(\alpha_{1}+\alpha_{r}+1)!}{(\alpha_{1}+1)!\alpha_{r}!}}.$

Therefore

$\frac{f(\alpha_{1}+1)}{f(\alpha_{1})}=\frac{2^{\alpha_{1}+k+1}-1}{2^{\alpha_{1}+1}-1}\frac{\alpha_{1}+1}{\alpha_{1}+\alpha_{r}+1}.$

In order to prove $f(\alpha_{1}+1)>f(\alpha_{1})$ we have to prove that

$\frac{2^{\alpha_{1}+k+1}-1}{2^{\alpha_{1}+1}-1}>\frac{\alpha_{1}+\alpha_{r}+1}{\alpha_{1}+1}.$

As $\frac{2^{\alpha_{1}+1+k}-1}{2^{\alpha_{1}+1}-1}>2^{k}$ and
$\alpha_{r}+1\geq\frac{\alpha_{1}+1+\alpha_{r}}{\alpha_{1}+1}$. So it is
sufficient enough to prove

$2^{k}>\alpha_{r}+1\Longleftrightarrow k>\log_{2}(\alpha_{r}+1).$

Now $\log_{2}t=k+f$, where $0<f<1$. So $k=\log_{2}t-f$. Therefore, we have to
prove that

$\log_{2}t-f>\log_{2}(\alpha_{r}+1)\Longleftrightarrow\log_{2}(\frac{t}{\alpha_{r}+1})>f.$

As $0<f<1$ it is sufficient enough to prove
$\log_{2}(\frac{t}{\alpha_{r}+1})>1$ since $\log x$ is a monotone increasing
function for $x>0$, i.e.

$\frac{t}{\alpha_{r}+1}>2\Longleftrightarrow p^{\alpha_{r}}>2\alpha_{r}+2$

for all prime $p\geq 5$ with $\alpha_{r}\geq 1$ and $p=3$ with $\alpha_{r}\geq
2$, which can easily be proved using mathematical induction. Hence
$\frac{X}{Y}$ is a monotone increasing function of $\alpha_{1}$. ∎

######  Theorem 4.2.

Among all $p$ groups of order $\leq n$, where $n\geq 4$ is a positive integer;
the elementary abelian group of order $2^{\alpha}$ where
$\alpha=\left[\log_{2}n\right]$, has the highest number of composition series.

###### Proof.

We know that among all finite $p$-groups of equal order the elementary abelian
group has the highest number of composition series. Let $G_{2}$ be an
elementary abelian group of order $2^{\alpha}$, where
$\alpha=\left[\log_{2}n\right]$ and let $G_{p}$ be an elementary abelian $p$
group of order $p^{\beta}$, where $\beta=\left[\log_{p}n\right]$ with $p\geq
3$ be a prime and $n\geq 4$. Now we know that if $\left[\log_{q}r\right]=t$
then $\left[\log_{q}qr\right]=t+1$ for all prime $q\geq 2$ and
$r,t\in\mathbb{N}$ with $q\leq r$. Obviously $2$ divides $p-1$. Therefore we
get

$2^{\left[\log_{2}n\right]}-1>\frac{p^{\left[\log_{p}n\right]}-1}{p-1},\leavevmode\nobreak\
\mbox{whenever}\leavevmode\nobreak\ n\geq 4.$ (1)

$\mid C_{G_{2}}\mid=\prod_{i=1}^{\left[\log_{2}n\right]}\left(2^{i}-1\right),$
$\mid
C_{G_{p}}\mid=\prod_{i=1}^{\left[\log_{p}n\right]}\left(\frac{p^{i}-1}{p-1}\right).$

From (1) it implies that

$\prod_{i=1}^{\left[\log_{2}n\right]}\left(2^{i}-1\right)>\prod_{i=1}^{\left[\log_{p}n\right]}\left(\frac{p^{i}-1}{p-1}\right)$

i.e.$\mid C_{G_{2}}\mid>\mid C_{G_{p}}\mid$. Therefore among all $p$ groups of
order $\leq n$, the elementary abelian group of order $2^{\alpha}$ has the
highest number of composition series where $\alpha=\left[\log_{2}n\right]$. ∎

######  Theorem 4.3.

Let $n\geq 4$ be a positive integer. Among all finite groups of order $\leq
n$, the elementary abelian group of order $2^{\alpha}$ has the highest number
of composition series where $\alpha=\left[\log_{2}n\right]$.

###### Proof.

$n\geq 4$ is a natural number and $\alpha=\left[\log_{2}n\right]$ and let $G$
be the elementary abelian group of order $2^{\alpha}$ then

$\mid C_{G}\mid=\prod_{i=1}^{\left[\log_{2}n\right]}\left(2^{i}-1\right).$

As we know that among all finite groups of order $n\geq 2$ , the abelian group
with elementary abelian Sylow subgroups has the highest number of composition
series. So in order to prove our claim we have to prove that

$\prod_{i=1}^{\left[\log_{2}n\right]}\left(2^{i}-1\right)\geq\prod_{i=1}^{q}\left(\prod_{j=1}^{\beta_{i}}\frac{p_{i}^{j}-1}{p_{i}-1}\right)\frac{\left(\sum_{i=1}^{q}\beta_{i}\right)!}{\prod_{i=1}^{q}\beta_{i}!}$

where each $p_{i}$ is a prime and $\beta_{i},q\in\mathbb{N}$ such that
$\prod_{i=1}^{q}p_{i}^{\beta_{i}}\leq n$ and the equality is hold if and only
if $q=1,p_{1}=2$ and $\beta_{1}=\left[\log_{2}n\right]$. The inequality will
be proved in the following steps. Step 1 : Let $H$ be an abelian group of
order $m$ with elementary abelian Sylow subgroups such that $4\leq m\leq n$
where $m=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\dots p_{r}^{\alpha_{r}}$ also
$\mid C_{H}\mid$denotes the number of distnct composition series of $H$. Let
$p_{r}$ be a prime such that $p_{r}\neq 2$ and
$p_{r}^{\alpha_{r}}=t,\left[\log_{2}t\right]=k$. Then
$2^{k}<p_{r}^{\alpha_{r}}.$ Let $H^{\prime}$ be the abelian group of order
$\frac{\mid H\mid.2^{k}}{p_{r}^{\alpha_{r}}}$ with elementary abelian Sylow
subgroups. Then $\mid H^{\prime}\mid<\mid H\mid$. We will prove that $\mid
C_{H^{\prime}}\mid>\mid C_{H}$$\mid$.

Step 2 : If $\mid H^{\prime}\mid=2^{k}p_{1}^{\alpha_{1}}\dots
p_{r-1}^{\alpha_{r-1}}$ i.e. $2$ is not a divisor of $m$ then we have
$k>\alpha_{r}$ and

$\prod_{i=1}^{k}\left(2^{k}-1\right)>\prod_{i=1}^{\alpha_{r}}\left(\frac{p_{r}^{i}-1}{p_{r}-1}\right).$

Therefore, obviously $\mid C_{H^{\prime}}\mid>\mid C_{H}\mid$. Step 3 : If
$\mid H\mid=2^{\alpha_{1}}p_{2}^{\alpha_{2}}\dots p_{r}^{\alpha_{r}}$ then
$\mid H^{\prime}\mid=2^{\alpha_{1}+k}p_{2}^{\alpha_{2}}\dots
p_{r-1}^{\alpha_{r-1}}.$ In this case we will prove that $\mid
C_{H^{\prime}}\mid>\mid C_{H}\mid$.

$\mid
C_{H}\mid=\prod_{i=1}^{\alpha_{1}}\left(2^{i}-1\right)\prod_{i=1}^{\alpha_{r}}\left(\frac{p_{r}^{\alpha_{r}}-1}{p_{r}-1}\right)\prod_{i=2}^{r-1}\left(\prod_{j=1}^{\alpha_{i}}\frac{p_{i}^{j}-1}{p_{i}-1}\right)\frac{\left(\sum_{i=1}^{r}\alpha_{i}\right)!}{\prod_{i=1}^{r}\alpha_{i}!},$

$\mid
C_{H^{\prime}}\mid=\prod_{i=1}^{\alpha_{1}+k}\left(2^{i}-1\right)\prod_{i=2}^{r-1}\left(\prod_{j=1}^{\alpha_{i}}\frac{p_{i}^{j}-1}{p_{i}-1}\right)\frac{\left(\sum_{i=1}^{r-1}\alpha_{i}+k\right)!}{\prod_{i=2}^{r-1}\alpha_{i}!\left(\alpha_{1}+k\right)!}.$

Therefore,

$\frac{\mid C_{H^{\prime}}\mid}{\mid
C_{H}\mid}=\frac{\prod_{i=\alpha_{1}+1}^{\alpha_{1}+k}\left(2^{i}-1\right)\left(k+s\right)!\alpha_{1}!\alpha_{r}!}{\prod_{j=1}^{\alpha_{r}}\left(\frac{p_{r}^{j}-1}{p_{r}-1}\right)\left(\alpha_{1}+k\right)!\left(\alpha_{r}+s\right)!},$

where $s=\sum_{i=1}^{r-1}\alpha_{i}.$ So in order to prove our claim we have
to prove that

$\prod_{i=\alpha_{1}+1}^{\alpha_{1}+k}\left(2^{i}-1\right)\left(k+s\right)!\alpha_{1}!\alpha_{r}!>\prod_{j=1}^{\alpha_{r}}\left(\frac{p_{r}^{j}-1}{p_{r}-1}\right)\left(\alpha_{1}+k\right)!\left(\alpha_{r}+s\right)!$
(2)

Note that $s=\sum_{i=1}^{r-1}\alpha_{i}\geq\alpha_{1},\Longrightarrow
s=\alpha_{1}+a,$ where $a\geq 0,\leavevmode\nobreak\
k>\alpha_{r}\Longrightarrow k=\alpha_{r}+b,\leavevmode\nobreak\
b>0;\leavevmode\nobreak\ a,b\in\mathbb{\mathbb{Z}}$. Then inequality ( 2)
reduces to

$\prod_{i=\alpha_{1}+1}^{\alpha_{1}+k}\left(2^{i}-1\right)\left(\alpha_{1}+\alpha_{r}+a+b\right)!\alpha_{1}!\alpha_{r}!>\prod_{j=1}^{\alpha_{r}}\left(\frac{p_{r}^{j}-1}{p_{r}-1}\right)\left(\alpha_{1}+\alpha_{r}+a\right)!\left(\alpha_{1}+\alpha_{r}+b\right)!$
(3)

[Here we are ignoring the case $p_{r}=3$ with $\alpha_{r}=1$, as $m\geq 4$ at
least one other case must exist.] Note

$\frac{\left(\alpha_{1}+\alpha_{r}+a+b\right)!\alpha_{1}!\alpha_{r}!}{\left(\alpha_{1}+\alpha_{r}+a\right)!\left(\alpha_{1}+\alpha_{r}+b\right)!}$

is a monotone increasing function of $a$ for fixed $\alpha_{1},\alpha_{r},b$.
So it is sufficient enough to prove the inequality for $a=0$ i.e.

$\prod_{i=\alpha_{1}+1}^{\alpha_{1}+k}\left(2^{i}-1\right)\alpha_{1}!\alpha_{r}!>\prod_{j=1}^{\alpha_{r}}\left(\frac{p_{r}^{j}-1}{p_{r}-1}\right)\left(\alpha_{1}+\alpha_{r}\right)!$
(4)

Let $X=\prod_{i=\alpha_{1}+1}^{\alpha_{1}+k}(2^{i}-1)\alpha_{1}!\alpha_{r}!$
and
$Y=\prod_{j=1}^{\alpha_{r}}(\frac{p_{r}^{j}-1}{p_{r}-1})(\alpha_{1}+\alpha_{r})!$.
Then according to lemma 4.1 $\frac{X}{Y}$ is a monotone increasing function of
$\alpha_{1}$ whenever $p_{r}$ and $\alpha_{r}$ are fixed. So it is
sufficiently enough to prove the inequality (4) for $\alpha_{1}=0$ i.e.

$\prod_{i=1}^{k}\left(2^{i}-1\right)>\prod_{j=1}^{\alpha_{r}}\left(\frac{p_{r}^{j}-1}{p_{r}-1}\right),$

which is following from theorem 4.2. As a result we prove that $\mid
C_{H^{\prime}}\mid>\mid C_{H}\mid$. Now we will repeat the same process in the
group $H^{\prime}$ and get a group $H^{\prime\prime}$ such that $\mid
C_{H^{\prime\prime}}\mid>\mid C_{H^{\prime}}\mid$. Since $\mid H\mid$is finite
then $\mid H\mid$has only finitely many prime divisors. So after a finite
number of steps we get an elementary abelian group of order $2^{\eta}$, where
$\eta\leq\left[\log_{2}n\right]$. Step 4 : Let $\mid
H\mid=2^{\alpha_{1}}3^{1}$ for any $\alpha_{1}\in\mathbb{N}$. Then $\mid
C_{H}\mid=\prod_{i=1}^{\alpha_{1}}\left(2^{i}-1\right)\left(\alpha_{1}+1\right)$.
Then we can find an elementary abelian group $H^{\prime}$ such that$\mid
H^{\prime}\mid=2^{\alpha_{1}+1}$ and $\mid
C_{H^{\prime}}\mid=\prod_{i=1}^{\alpha_{1}+1}\left(2^{i}-1\right)$. Obviously
$\mid C_{H^{\prime}}\mid>\mid C_{H}\mid$ as $2^{\alpha_{1}+1}>\alpha_{1}+2$
for each $\alpha_{1}\in\mathbb{N}$. Among all $2$ groups of order $\leq n$,
$G$ has the highest number of composition series which proves the theorem. ∎

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