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#  Small Representations of Permutation Groups by Embedding the Domain in an
Ordered Set

Bernd S. W. Schröder
School of Mathematics and Natural Sciences
The University of Southern Mississippi
118 College Avenue, #5043
Hattiesburg, MS 39406

###### Abstract

We present a representation for permutation groups as the automorphism group
of an ordered set $U$ such that the automorphism group’s action on a subset
$T\subseteq U$ is the permutation group itself. For many imprimitive
permutation groups, this representation uses fewer points than the smallest
representations to date.

AMS subject classification (2020): 06A07, 06A06, 20C30, 20B35, 20B25
Key words: Ordered set; automorphism; permutation group; representation

##  1 Introduction

König’s ([14]) classical question when a given group can be represented as the
automorphism group of a (finite) graph can be considered in an abstract and a
concrete version. The abstract version asks if, for a group $G$, there is a
graph (or another combinatorial structure) whose automorphism group (with no
further structural demands made) is isomorphic to $G$. This question was
answered affirmatively for graphs and all finite groups in [7, 8, 15]. For
ordered sets and finite groups, the question was answered in [4, 19, 2] with
successively smaller underlying ordered sets by refining Birkhoff’s idea from
[4]. Interestingly, in [16], orthomodular lattices with a given automorphism
group are constructed by using Frucht’s construction for graphs (by way of
[15]), not Birkhoff’s construction for ordered sets. The smallest known
representations via ordered sets have $3|G|$ elements, see [1], Corollary 4.3.

The concrete version of König’s question asks whether a given permutation
group on $\\{1,\ldots,n\\}$ can be the automorphism group of a graph (or
another combinatorial structure) whose vertex set is $\\{1,\ldots,n\\}$. The
answer to this question is negative for graphs (see [13]) as well as for
directed graphs (see [12]). This negative answer led to much work to determine
which permutation groups can be represented in this fashion, with [3, 10, 11]
being recent works that also provide an overview of further research. A
permutation group is regular, iff no nontrivial permutation has a fixed point.
In particular, the standard representation of a group as a permutation group
on itself is regular. For regular permutation groups, the concrete
representation question was settled for graphs in [9], and for directed graphs
in [1]. For hypergraphs, it was addressed and “essentially settled” in [5, 6].

When analyzing (the number of) automorphisms of an ordered set, it is natural
to wonder if every permutation group can occur as the action of an
automorphism group on an orbit. Indeed, the main idea for our construction was
conceived in this context in [18]. For permutation groups with a concrete
hypergraph representation, the answer is positive by identifying the
hypergraph with the ordered set that consists of the hyperedges and of the
singleton sets for the vertices, ordered by containment. The question remains
whether, for every permutation group $G$ on a domain $D$, there is a
representation with an ordered set $U_{G}$ that contains $D$ such that $G$ is
isomorphic to the automorphism group ${\rm Aut}(U_{G})$ and such that the
action ${\rm Aut}(U_{G})\circlearrowleft D$ of the automorphism group ${\rm
Aut}(U_{G})$ on the subset $D\subseteq U_{G}$ is $G$. Such a representation is
given in Section 2.

Recall that a block for a permutation group $G$ on $\\{1,\ldots,n\\}$ is a set
$B\subseteq\\{1,\ldots,n\\}$ such that, for all $\Phi\in G$, we have that
$\Phi[B]\cap B\not=\emptyset$ implies $\Phi[B]=B$. Recall that a permutation
group on $\\{1,\ldots,n\\}$ is called primitive iff the singletons and
$\\{1,\ldots,n\\}$ are its only blocks. We will see in Section 3 that the
construction’s explicit use of an underlying block structure, for many
imprimitive permutation groups, generates a representation with fewer than
$3|G|$ elements.

##  2 The Construction

Recall that an ordered set is a set $P$ equipped with a reflexive,
antisymmetric and transitive relation $\leq$, the order relation, and that the
strict comparability $<$ is $\leq\setminus=$. [Note: Any inadvertently
unexplained order terminology is consistent with [17].] Recall that an
antichain is a subset of an ordered set such that no two distinct elements are
comparable and that, when an element $p$ is less than every element of a set
$X$, we write $p<X$. Let $G\subseteq S_{n}$ be a permutation group. For
$A\subseteq\\{1,\ldots,n\\}$, we set $G\cdot A=\\{\sigma[A]:\sigma\in G\\}$.
Recall that the orbit of an element $x$ is the set $G\cdot
x=\\{\sigma(x):\sigma\in G\\}$, and that the orbits partition
$\\{1,\ldots,n\\}$.

###### Definition 2.1

Let $G$ be a subgroup of the symmetric group $S_{n}$ and let
$\\{O_{1},\ldots,O_{\ell}\\}$ be the partition of $\\{1,\ldots,n\\}$ into the
orbits of $G$. For every $j\in\\{1,\ldots,\ell\\}$, let $B_{j}\subseteq O_{j}$
be a block of $G$. We will call the partition ${\cal
B}:=\bigcup_{j=1}^{\ell}G\cdot B_{j}$ of $\\{1,\ldots,n\\}$ a block partition
with orbit cut $\\{B_{1},\ldots,B_{\ell}\\}$. For every
$j\in\\{1,\ldots,\ell\\}$, we let $m_{j}:=|G\cdot B_{j}|$. For all $B\in{\cal
B}$, we define $G|_{B}:=\left\\{\theta|_{B}:\theta\in G\right\\}$,

Note that the $\mu\in G|_{B}$ can map $B$ to any $C\in G\cdot B$.

###### Definition 2.2

Let $G\subseteq S_{n}$ be a subgroup of the symmetric group $S_{n}$ on
$\\{1,\ldots,n\\}=:T$, let ${\cal B}$ be a block partition for $G$, and, for
all $\mu\in\bigcup_{B\in{\cal B}}G|_{B}$, let $D^{\mu}:={\rm
dom}(\mu)\times\\{\mu\\}$. Moreover, let
$F:=\\{\ell_{1},u_{1},\ell_{2},u_{2},\ldots,\ell_{n},u_{n}\\}$. We define
$U_{G\looparrowright{\cal B}}:=F\cup
G\cup\left(\bigcup_{\mu\in\bigcup_{B\in{\cal B}}G|_{B}}D^{\mu}\right)\cup T$
to be the ordered set with the following comparabilities.

  1. 1.

$\ell_{1}<u_{1}>\ell_{2}<u_{2}>\cdots>\ell_{n}<u_{n}$, that is, $F$ is a fence
with $2n$ elements.

  2. 2.

Each of the sets $G$, $\bigcup_{\mu\in\bigcup_{B\in{\cal B}}G|_{B}}D^{\mu}$,
and $T$ is an antichain.

  3. 3.

For every $\mu\in\bigcup_{B\in{\cal B}}G|_{B}$ and every $j\in{\rm dom}(\mu)$,
we let $(j,\mu)>u_{j}\in F$ and we let $(j,\mu)<\mu(j)\in T$.

  4. 4.

For every $\theta\in G$, we let $\theta<\bigcup_{B\in{\cal
B}}D^{\theta|_{B}}$.

  5. 5.

Plus all comparabilities dictated by transitivity.

###### Example 2.3

The simplest visualization for ordered sets $U_{G\looparrowright{\cal B}}$ is
obtained for the trivial block partition ${\cal B}=\\{\\{1,\ldots,n\\}\\}$. In
this case, for every $\theta\in G$, we have that
$D^{\theta}=D^{\theta|_{\\{1,\ldots,n\\}}}$ is an antichain with $n$ elements,
and $\theta$ is simply a lower bound of $D^{\theta}$.

Figure 1 shows the resulting ordered set for the symmetric group $S_{3}$. The
sets $D^{\theta}$ are indicated with dashed boxes. As the proof of Theorem 2.4
below will show, every permutation $\sigma$ of $T$ is induced by an
automorphism that fixes every point in $F$: Every $\theta$ is mapped to
$\sigma\theta$, and $D^{\theta}$ is mapped $D^{\sigma\theta}$, with
$(j,\theta)\mapsto(j,\sigma\theta)$. Moreover, there are no further
automorphisms.

Ordered sets $U_{S_{n}\looparrowright\\{\\{1,\ldots,n\\}\\}}$ are obtained
similarly. For subgroups $G$ of $S_{n}$, ordered sets
$U_{G\looparrowright\\{\\{1,\ldots,n\\}\\}}$ are obtained by, for all
$\theta\not\in G$, deleting $\theta$ and $D^{\theta}$ from
$U_{S_{n}\looparrowright\\{\\{1,\ldots,n\\}\\}}$.

$D^{\rm id}$${\rm
id}$$D^{(23)}$${(23)}$$D^{(13)}$${(13)}$$D^{(12)}$${(12)}$$D^{(123)}$${(123)}$$D^{(132)}$${(132)}$$T$$F$

Figure 1:  Hasse diagram of the ordered set
$U_{S_{3}\looparrowright\\{\\{1,2,3\\}\\}}$.

Recall that $x$ is a lower cover of $y$ and $y$ is an upper cover of $x$ iff
$x<y$ and there is no $z\in P\setminus\\{x,y\\}$ such that $x<z<y$. Recall
that a minimal element is an element $b$ such that there is no $x$ with $x<b$,
that a maximal element is an element $u$ such that there is no $x$ with $x>u$,
that a chain is an ordered set such that any two elements are comparable, that
the length of a chain is one less than its number of elements, and that the
rank of an element $x$ is the length of a longest chain from $x$ to a minimal
element.

###### Theorem 2.4

Let $G$ be a subgroup of $S_{n}$ and let ${\cal B}$ be a block partition for
$G$. The automorphism group ${\rm Aut}\left(U_{G\looparrowright{\cal
B}}\right)$ is isomorphic to $G$ via the restriction $\Psi\mapsto\Psi|_{T}$.

Proof. Clearly, because $(\Phi\Psi)|_{T}=\Phi|_{T}\Psi|_{T}$, the function
$\Psi\mapsto\Psi|_{T}$ is a group homomorphism.

Claim. Every $\Psi\in{\rm Aut}\left(U_{G\looparrowright{\cal B}}\right)$ is
uniquely determined by $\sigma:=\Psi|_{T}$, which is an element of $G$.

Let $\Psi\in{\rm Aut}\left(U_{G\looparrowright{\cal B}}\right)$ and let
$\sigma:=\Psi|_{T}$. Because automorphisms map maximal elements to maximal
elements, $\Psi$ maps $T$ to itself. Because automorphisms preserve the
lengths of chains and map minimal elements to minimal elements, $\Psi$ maps
each of the sets $\\{\ell_{1},\ell_{2},\ldots,\ell_{n}\\}$ and $G$ to itself.
Because automorphisms preserve the rank, $\Psi$ maps each of
$\\{u_{1},u_{2},\ldots,u_{n}\\}$ and $\bigcup_{\mu\in\bigcup_{B\in{\cal
B}}G|_{B}}D^{\mu}$ to itself. Hence $\Psi[F]=F$, and, because $F$ has exactly
one automorphism, $\Psi|_{F}={\rm id}_{F}$.

For the following, let $\mu\in\bigcup_{B\in{\cal B}}G|_{B}$. Clearly, there is
an $A\in{\cal B}$ such that $\mu\in G|_{A}$, that is, $A$ is the domain of
$\mu$.

First let $j\in{\rm dom}(\mu)=A$. Then there are a $\nu\in\bigcup_{B\in{\cal
B}}G|_{B}$ and an $i\in{\rm dom}(\nu)$ such that $\Psi(j,\mu)=(i,\nu)\in
D^{\nu}$. Now $u_{j}<(j,\mu)$ implies
$u_{j}=\Psi(u_{j})<\Psi\left(j,\mu\right)=(i,\nu)$. Because $(j,\nu)$ is the
unique upper cover of $u_{j}$ in $D^{\nu}$, we obtain $i=j$, that is
$\Psi\left(j,\mu\right)=(j,\nu)$. Because $j\in{\rm dom}(\nu)$ and ${\cal B}$
is a partition, by definition, we obtain that ${\rm dom}(\nu)=A$.

Next, let $(a,\mu),(b,\mu)\in D^{\mu}$ and consider $\Psi(a,\mu)=:(a,\nu_{a})$
and $\Psi(b,\mu)=:(b,\nu_{b})$. By definition, $(a,\mu)$ and $(b,\mu)$ have
the same lower covers in $G$. Hence $\Psi(a,\mu)=(a,\nu_{a})$ and
$\Psi(b,\mu)=(b,\nu_{b})$ have the same lower covers in $G$. Thus, for all
$\theta\in G$ with $\theta<\nu_{a},\nu_{b}$, we have that
$\nu_{a}=\theta|_{A}=\nu_{b}$. Because $(a,\mu),(b,\mu)\in D^{\mu}$ were
arbitrary, we conclude that there is a $\nu\in G|_{A}$ such that
$\Psi\left[D^{\mu}\right]=D^{\nu}$.

For every $j\in A$, we have that $u_{j}<(j,\mu)<\mu(j)$ implies
$u_{j}=\Psi(u_{j})<\Psi(j,\mu)=(j,\nu)=\Psi(j,\mu)<\Psi\left(\mu(j)\right)=\sigma\left(\mu(j)\right)$.
By definition, we must have $\nu=\sigma\mu$. Therefore
$\Psi\left[D^{\mu}\right]=D^{\sigma\mu}$, and, for all $(j,\mu)\in
A\times\\{\mu\\}$, we have $\Psi(j,\mu)=(j,\sigma\mu)$.

Because the above holds for all $\mu\in\bigcup_{B\in{\cal B}}G|_{B}$, for
every $\theta\in G$, we have that $\Psi(\theta)$ is the lower cover of
$\bigcup_{B\in{\cal B}}D^{\sigma\theta|_{B}}$. Consequently
$\Psi(\theta)=\sigma\theta$. On one hand, this concludes the proof that
$\sigma=\Psi|_{T}$ determines $\Psi(x)$ for every $x\in
U_{G\looparrowright{\cal B}}$. On the other hand, using any $\theta\in G$, we
have that $\sigma\theta=\Psi(\theta)\in G$ implies
$\sigma=\Psi(\theta)\theta^{-1}\in G$. This completes the proof of the Claim.
In particular, $\Psi\mapsto\Psi|_{T}$ is injective and it maps into $G$.

Finally, for surjectivity, let $\sigma\in G$. We define $\Phi_{\sigma}$ by
$\Phi_{\sigma}|_{F}:={\rm id}|_{F}$, by, for every $j\in T$, setting
$\Phi_{\sigma}(j):=\sigma(j)$, for every $\mu\in\bigcup_{B\in{\cal B}}G|_{B}$
and every $j\in{\rm dom}(\mu)$, setting
$\Phi_{\sigma}\left(j,\mu\right):=(j,\sigma\mu)$, and, for every $\theta\in
G$, setting $\Phi_{\sigma}(\theta)=\sigma\theta$. Clearly, $\Phi_{\sigma}$ is
injective, hence surjective, $\Phi_{\sigma}$ preserves order on $F$, and every
comparability $\theta<D^{\theta|_{B}}$ is mapped to
$\Phi_{\sigma}(\theta)=\sigma\theta<D^{\sigma\theta|_{B}}=\Phi_{\sigma}\left[D^{\theta|_{B}}\right]$.
Finally, for every $\mu\in\bigcup_{B\in{\cal B}}G|_{B}$ and every $j\in{\rm
dom}(\mu)$, we have that $u_{j}<(j,\mu)<\mu(j)$ implies
$\Phi_{\sigma}(u_{j})=u_{j}<(j,\sigma\mu)=\Phi_{\sigma}\left(j,\mu\right)=(j,\sigma\mu)<\sigma\left(\mu(j)\right)=\Phi_{\sigma}\left(\mu(j)\right)$.
Hence $\Phi_{\sigma}\in{\rm Aut}(U_{G\looparrowright{\cal B}})$ and
$\Psi\mapsto\Psi|_{T}$ is surjective.

Now that we have a representation, we turn to its size.

###### Definition 2.5

Let $G$ be a subgroup of $S_{n}$ and let ${\cal B}$ be a block partition for
$G$. For every $B\in{\cal B}$, we define $G\circlearrowleft
B:=\\{\sigma|_{B}:\sigma\in G,\sigma[B]=B\\}$. Moreover, we define
$G\circlearrowleft{\cal B}$ to be the group of permutations on ${\cal B}$
induced by $G$ by, for every $\sigma\in G$, mapping each block to its image
under $\sigma\in G$.

###### Proposition 2.6

Let $G$ be a subgroup of $S_{n}$, let ${\cal B}$ be a block partition for $G$,
and let $\left\\{B_{1},\ldots,B_{\ell}\right\\}\subseteq{\cal B}$ be an orbit
cut. Then

$\displaystyle{\left|U_{G\looparrowright{\cal
B}}\right|=|G|+3n+\sum_{i=1}^{\ell}|B_{i}|\cdot
m_{i}^{2}\cdot|G\circlearrowleft B_{i}|.}$

In case $G\circlearrowleft{\cal B}$ is transitive, with $N:=\left\\{\sigma\in
G:\sigma[B_{i}]=B_{i},i=1,\ldots,m\right\\}$, we obtain
$\displaystyle{\left|U_{G\looparrowright{\cal
B}}\right|=|G|\left(1+{3n\over|G|}+{|G\circlearrowleft B_{1}|\over|N|}{|{\cal
B}|n\over|G\circlearrowleft{\cal B}|}\right).}$ In case $G$ is transitive, we
obtain $\displaystyle{\left|U_{G\looparrowright{\cal
B}}\right|\leq|G|\left(1+{|G\circlearrowleft
B_{1}|\over|N|}\left(n+3\right)\right).}$

Proof. We have $|T|=n$ and $|F|=2n$.

Now let $i\in\\{1,\ldots,\ell\\}$. Then, because every $\mu\in G|_{B_{i}}$
maps onto exactly one $B\in G\cdot B_{i}$, we have
$\left|G|_{B_{i}}\right|=m_{i}\cdot\left|G\circlearrowleft B_{i}\right|$.
Moreover, for every $B\in G\cdot B_{i}$, and every $\mu\in G|_{B}$, we have
$\left|D^{\mu}\right|=|B_{i}|$. Therefore, because every $B\in{\cal B}$ is in
exactly one $G\cdot B_{i}$, we obtain $\left|\bigcup_{\mu\in\bigcup_{B\in{\cal
B}}G|_{B}}D^{\mu}\right|=\sum_{i=1}^{\ell}|B_{i}|\cdot
m_{i}\cdot\left|G|_{B_{i}}\right|=\sum_{i=1}^{\ell}|B_{i}|\cdot m_{i}\cdot
m_{i}\cdot|G\circlearrowleft B_{i}|$. Hence $\left|U_{G\looparrowright{\cal
B}}\right|=|G|+3n+\sum_{i=1}^{\ell}|B_{i}|\cdot
m_{i}^{2}\cdot|G\circlearrowleft B_{i}|$.

In case $G\circlearrowleft{\cal B}$ is transitive, we have $\ell=1$,
$m_{1}=|{\cal B}|$ and $m_{1}|B_{1}|=n$. Moreover, $N$ is a normal subgroup of
$G$ and $G/N$ is isomorphic to $G\circlearrowleft{\cal B}$, so
$|G|=|N|\cdot|G\circlearrowleft{\cal B}|$. Now

$\displaystyle\left|U_{G\looparrowright{\cal B}}\right|$ $\displaystyle=$
$\displaystyle|G|+3n+|B_{1}|\cdot m_{1}\cdot m_{1}\cdot|G\circlearrowleft
B_{1}|$ $\displaystyle=$ $\displaystyle|G|+3n+|{\cal B}|n|G\circlearrowleft
B_{1}|$ $\displaystyle=$ $\displaystyle|G|+|G|{3n+|{\cal
B}|n|G\circlearrowleft B_{1}|\over|G|}$ $\displaystyle=$
$\displaystyle|G|\left(1+{3n\over|G|}+{|G\circlearrowleft
B_{1}|\over|N|}{|{\cal B}|n\over|G\circlearrowleft{\cal B}|}\right).$

Finally, in case $G$ is transitive, we have $|G\circlearrowleft{\cal
B}|\geq|{\cal B}|$, $|G|\geq|N|\cdot|{\cal B}|$, and $|G\circlearrowleft
B_{1}|\geq|B_{1}|={n\over|{\cal B}|}$. Hence we obtain

${3n\over|G|}+{|G\circlearrowleft B_{1}|\over|N|}{|{\cal
B}|n\over|G\circlearrowleft{\cal B}|}\leq{3n\over|N||{\cal
B}|}+{|G\circlearrowleft B_{1}|\over|N|}{|{\cal B}|n\over|{\cal
B}|}\leq{|G\circlearrowleft B_{1}|\over|N|}\left(n+3\right)$

and the final inequality follows.

##  3 Discussion

To see that, for many imprimitive permutation groups, Theorem 2.4 gives a very
compact representation, consider the terms in Proposition 2.6. For many
permutation groups, we have $|G|\geq 3n$, and we often have $|G|\gg 3n$.
Moreover, ${|G\circlearrowleft B_{1}|\over|N|}\leq 1$ and, for imprimitive
$G$, we often have ${|G\circlearrowleft B_{1}|\over|N|}\ll 1$. Finally, for
transitive actions $G\circlearrowleft{\cal B}$, we have
$|G\circlearrowleft{\cal B}|\geq|{\cal B}|$ and often $|G\circlearrowleft{\cal
B}|\gg|{\cal B}|$. Consequently, a demand ${3n\over|G|}+{|G\circlearrowleft
B_{1}|\over|N|}{|{\cal B}|n\over|G\circlearrowleft{\cal B}|}<1$ is a
relatively mild condition, which, via Theorem 2.4 and Proposition 2.6, yields
a representation with at most $2|G|$ points. For transitive imprimitive
groups, with sufficiently many independent actions on the blocks $B\in{\cal
B}$, we obtain ${|G\circlearrowleft B_{1}|\over|N|}\left(n+3\right)<1$, and
again we have a representation with fewer than $2|G|$ points.

In fact, it is an easy task to construct sequences
$\\{G_{k}\\}_{k=1}^{\infty}$ of transitive imprimitive permutation groups
$G_{k}$ whose order goes to infinity and such that there are nontrivial block
partitions ${\cal B}_{k}$ such that the quotient
${\left|U_{G_{k}\looparrowright{\cal B}_{k}}\right|\over|G_{k}|}$ converges to
$1$. For example, if $G_{k}$ is the group on
$\left\\{1,\ldots,k^{2}\right\\}$, generated by the cycles $(1+\ell k,2+\ell
k,\ldots,k+\ell k)$, $\ell=0,\ldots,k-1$, plus any transitive action on the
block partition ${\cal B}_{k}:=\left\\{\left\\{1+\ell k,2+\ell k,\ldots,k+\ell
k\right\\}:\ell=0,\ldots,k-1\right\\}$, then $1+{|G\circlearrowleft
B_{1}|\over|N|}\left(n+3\right)\leq 1+{k\over k^{k}}\left(k^{2}+3\right)\to 1$
(indeed quite rapidly) as $k\to\infty$. This simple example indicates how
sufficiently many independent actions on the blocks lead to a rather compact
representation: Indeed, with the partition above and cyclic action on $B_{1}$,
$|N|\geq k^{4}$ would still suffice.

Finally, via the trivial block partition $\\{\\{1,\ldots,n\\}\\}$, Theorem 2.4
allows for a “semi-concrete” representation of any permutation group
$G\subseteq S_{n}$ via ${\rm
Aut}\left(U_{G\looparrowright\\{\\{1,\ldots,n\\}\\}}\right)$. However, by
Proposition 2.6,
$\left|U_{G\looparrowright\\{\\{1,\ldots,n\\}\\}}\right|=(n+1)|G|+3n>3|G|$. It
must also be noted that, when $n=|G|$, which is the case for the standard,
though combinatorially voluminous, representation of an abstract group as a
permutation group on itself, $U_{G\looparrowright\\{\\{1,\ldots,n\\}\\}}$
makes for a rather large representation. On the positive side,
$U_{G\looparrowright\\{\\{1,\ldots,n\\}\\}}$ can easily be turned into a
lattice with $(n+1)|G|+3n+3$ elements and the same property as described in
Theorem 2.4 by adding a top element ${\bf 1}$, a bottom element ${\bf 0}$, and
a “center element” $c$ such that $F<c<T$, and, for all $\theta\in G$, we have
$\theta<c<T$.

Acknowledgment. The author very much thanks Frank a Campo for reading an
earlier version of this manuscript and for identifying a way to eliminate an
unnecessary technical condition in the construction.

##  4 Declarations

  * •

Ethical Approval. Not applicable, because no human or animal studies were
involved.

  * •

Competing Interests. The author has no relevant financial or non-financial
interests to disclose.

  * •

Authors’ Contributions. This is a single author paper with all work done by
the author. Contributions by non-authors are acknowledged above.

  * •

Funding. No funds, grants, or other support was received.

  * •

Availability of Data and Materials. Data sharing is not applicable, as no
datasets were generated or analysed herein.

## References

  * [1] L. Babai, Finite digraphs with given regular automorphism groups, Period. Math. Hungar. 11 (4) (1980), 257–270
  * [2] J. A. Barmak, E. G. Minian, Automorphism Groups Of Finite Posets, Discrete Mathematics 309 (2009), 3424–3426
  * [3] J. A. Barmak(2023), Regular and semi-regular representations of groups by posets, available at https://arxiv.org/abs/2307.03106
  * [4] G. Birkhoff, On groups of automorphisms, Rev. Un. Mat. Argentina 11 (1946), 155–157
  * [5] S. Foldes, N. Singhi, Regular representation of abelian groups by 3-uniform hypergraphs, Ars Combin. 3 (1977), 15–20
  * [6] S. Foldes, N. Singhi, Regular Representation of Finite Groups by Hypergraphs, Canadian Journal of Mathematics 30(5) (1978), 946–960
  * [7] R. Frucht, Herstellung von Graphen mit vorgegebener abstrakter Gruppe, Compositio Mathematica tome 6 (1939), 239–250
  * [8] R. Frucht, Graphs of degree three with a given abstract group, Canad. J. Math. 1 (1949), 365–378
  * [9] C. Godsil, GRRs for nonsolvable groups, Algebraic methods in graph theory. Colloq. Math. Soc. János Bolyai, Amsterdam, New York (1981), 221–239
  * [10] M. Grech, A. Kisielewicz, Abelian permutation groups with graphical representations, Journal of Algebraic Combinatorics 55 (2022), 513–531
  * [11] M. Grech, A. Kisielewicz, Orbit closed permutation groups, relation groups, and simple groups, Journal of Algebraic Combinatorics 57 (2023), 1045–1072
  * [12] R. Hemminger, On the group of a directed graph, Canadian Journal of Mathematics 18 (1966), 211–220
  * [13] I. N. Kagno, Linear Graphs of Degree $\leq 6$ and Their Groups, Amer. J. Math. 68 (1946), 505–520, with corrections added in American Journal of Mathematics, Vol. 69, No. 4 (Oct., 1947), p. 872 and American Journal of Mathematics, Vol. 77, No. 2 (Apr., 1955), p. 392
  * [14] D. König, Theorie der endlichen und unendlichen Graphen, Kombinatorische Topologie der Streckenkomplexe, (Leipzig), 1936
  * [15] G. Sabidussi, Graphs with given group and given graph-theoretical properties, Canadian Journal of Mathematics, Volume 9 515–525 (1957),
  * [16] G. Schrag, Every finite group is the automorphism group of some finite orthomodular lattice, Proc. Amer. Math. Soc. 55 (1976), 243–249
  * [17] B. Schröder, Ordered Sets – An Introduction with Connections from Combinatorics to Topology (second edition), Birkhäuser Verlag, Boston, Basel, Berlin, 2016
  * [18] B. Schröder (2023), Towards the Automorphism Conjecture I: Combinatorial Control, submitted to Order, available at https://arxiv.org/abs/2308.05715
  * [19] M.C. Thornton, Spaces with given homeomorphism groups, Proc. Amer. Math. Soc. 33 (1972), 127–131