Some time ago, I was discussing the Groups, Rings and Modules course with a friend, and this inspired me to conjecture and investigate a result in group theory.

The text is accessible to any reader with a sound grasp of standard group theory results (e.g. group actions, classification of finite abelian groups). The proofs are explained in particular detail, for the purpose of a not too taxing, enjoyable read!

Introduction

Groups encapsulate the idea of symmetry, and they hence arise in a plethora of mathematical topics.

Indeed, Cayley’s Theorem (see below) shows that every finite group is isomorphic to a subgroup of a symmetric group – that is to say, given a group $G$ of order $n$ , we can think of $G$ as a set of symmetries on $n$ objects.

But now a natural question to ask: do we really need all $n$ objects? Indeed, $D_6$ , the dihedral group of $6$ elements is in fact isomorphic to $S_3$ – it is the set of symmetries of just three objects, let alone six. But $C_2\times C_2$ is not a subgroup of $S_3$ , thus it can only be embedded in $S_4$ . Which groups share this property – namely where we need all of our $n$ objects?

We define such a group as a Bountiful Group, and explore their properties. In fact, our final result shows that it is possible to classify them all.

Notation and Preliminaries

Cayley’s Theorem. Let $G$ be a group of order $n$ . Then $G$ is isomorphic to a subgroup of $S_n$ .

Cauchy’s Theorem. If $p$ divides the order of a (finite) group $G$ say, then $G$ has an element of order $p$ .

Notation:

$S_n$ denotes the symmetric group of $n$ elements. A finite $p$ -group is a group of order a power of $p$ .
The order of a group $\left| G \right|$ is the number of elements in it. The order of $g \in G$ is the smallest positive integer $k$ such that $g^k = e$ (if it exists). $S_n$ denotes the symmetric group of $n$ elements. A finite $p$ -group is a group of order a power of $p$ .
$G/H$ denotes the set of left cosets of a subgroup $H$ in a group $G$ .

Classification

Definition. A Bountiful Group is a (finite) group of order $n$ that is not isomorphic to a subgroup of $S_k$ for all $k<n$ .

In fact, it is enough that it is not isomorphic to a subgroup of $S_{n-1}$ since $k < l \implies S_k \leq S_l$ , but we choose the above definition as it reflects the motivation of symmetry given in the introduction.

We first note a useful result, and a standard result.

Claim 1. $S_m \times S_n$ is isomorphic to a subgroup of $S_{m+n}$ , where $m,n$ are positive integers.

Proof. Consider $m+n$ objects, and an element $(g_m, g_n)$ of $S_m \times S_n$ . Let $g_m$ act on the first $m$ objects and $g_n$ act on the other $n$ objects. This corresponds to a unique element of $S_{m+n}$ , giving an injective homomorphism $\phi$ say. Thus $S_m \times S_n \cong \mathrm{Im} (\phi)$ by the first isomorphism theorem. $\blacksquare$

Claim 2. If all elements of a group $G$ have order two, then $G$ is abelian.

Proof. $xy = (xy)^{-1} = y^{-1}x^{-1} = yx. \ \blacksquare$

We now find a restriction on possible orders of bountiful groups.

Lemma. Let $G$ be a bountiful Group. Then $G$ has order a prime power.

Proof. Suppose not. Let $G$ be a bountiful group of order $n$ . Then there exist two distinct primes $p,q$ that both divide $n$ . By Cauchy’s Theorem, there exist two elements $h,k \in G$ with orders $p,q$ respectively.
Let $H$ denote the cyclic group (of order $p$ ) generated by $h$ , and $K$ similarly for $k$ .
The left regular action of $G$ on $G/H$ induces a homomorphism $\phi_p : G \rightarrow \mathrm{Sym}(G/H)$ . We get $\phi_q : G \rightarrow \mathrm{Sym}(G/K)$ similarly.

Now consider the combined left regular action $*$ of $G$ on $G/H \times G/K$ via:

$\begin{aligned} * \ : \ G \times (G/H \times G/K) & \rightarrow G/H \times G/K \\ g * (g_p H, g_q K) & = (gg_p H, gg_q K) \end{aligned}$ .

This induces a homomorphism $\phi$ as follows:

$\begin{aligned} \phi \ : \ G & \rightarrow \mathrm{Sym}(G/H) \times \mathrm{Sym}(G/K) \\ g & \mapsto (\phi_p(g), \phi_q(g))\end{aligned}$

Crucially, this is stronger than the natural homomorphism to $\mathrm{Sym}(G/H \times G/K)$ .

Consider the kernel of $\phi$ . Suppose $g \in \ker(\phi)$ . Then $g \in H, g \in K$ by construction, i.e. $g \in H \cap K$ . But $H \cap K = \{e\}$ due to the coprime orders of $H$ and $K$ . Hence the kernel is trivial. Thus the homomorphism is injective.

By the first isomorphism theorem, $G$ is isomorphic to a subgroup of $\mathrm{Sym}(G/H) \times \mathrm{Sym}(G/K)$ . But by Claim 1, $\mathrm{Sym}(G/H) \times \mathrm{Sym}(G/K) \leq \mathrm{Sym}(\left| G/H\right| + \left| G/K \right|) = \mathrm{Sym}(\frac{n}{p} + \frac{n}{q})$ .

Since $p,q$ are distinct primes, so $\frac{n}{p} + \frac{n}{q} < n$ (smallest pair $p,q$ is $2,3$ ). $G \leq \mathrm{Sym}(\frac{n}{p} + \frac{n}{q})$ , so $G$ is not a bountiful group, contradiction. $\blacksquare$

We can now prove the main result.

Theorem. Group $G$ is a bountiful group if and only if it is isomorphic to $C_2 \times C_2$ or a $p$ -group with a unique subgroup of order $p$ .

Proof.

Forward Direction.
Let $G$ be a bountiful group. The lemma from above gives that $G$ is a $p$ -group for some prime, order $p^k$ .

Case $p$ odd: Suppose for contradiction that $G$ does not have a unique subgroup of order $p$ . By Cauchy’s Theorem, there exists at least one subgroup of order $p$ . Thus we can find two non-equal subgroups $H, K$ of order $p$ in $G$ . Then $H \cap K = \{e\}$ , since if $H$ and $K$ shared a non-identity element, then this element would generate both groups, giving $H = K$ .

We now recycle the argument used in the lemma above using $H$ and $K$ . The kernel is trivial by construction of $H \cap K = \{e\}$ . Thus $G \leq \mathrm{Sym}(\frac{n}{p} + \frac{n}{p})$ , but $\frac{n}{p} + \frac{n}{p} = 2p^{k-1} < p^k$ since $p$ odd, so $G$ is not a bountiful group, contradiction.

Case $p=2$ : The argument above fails, for the inequality at the end is not strict. So we need something slightly different.

If all the elements in $G$ have order 2, then $G$ is abelian by Claim 2 above. Thus by the classification of abelian groups, $G \cong C_2 \times C_2 \times \dots \times C_2$ . But by extending the idea of Claim 1, $G \cong S_2 \times S_2 \times \dots \times S_2 \leq S_{2k}$ .

If $k>2$ , then $2k < 2^k$ , so $G$ is not a bountiful group, contradiction. Else either $k=1$ , and $C_2$ indeed has a unique subgroup of order 2, or $k=2$ , yielding $C_2\times C_2$ .

Backward Direction.

$C_2 \times C_2$ is not isomorphic to a subgroup of $S_3$ by orders (4 does not divide 6), so $C_2 \times C_2$ is a bountiful group.
Now let $G$ be a group of order $n = p^k$ , with a unique subgroup of order $p$ , $P$ say.

Claim. Every non-trivial subgroup of $G$ contains $P$ .

Proof. Let $H$ be a non-trivial subgroup of $G$ . Then $p$ divides $\left| H \right|$ , so by Cauchy’s Theorem $H$ contains an element of order $p$ . But then this element generates a subgroup of order $p$ in $H$ , which is also a subgroup of order $p$ in $G$ . So this subgroup is $P$ , hence $P \leq H$ . $\blacksquare$

Now suppose $G$ is isomorphic to a subgroup of $S_{n-1}$ for contradiction. That is to say, there is an injective homomorphism $\phi : G \rightarrow S_{n-1}$ . This then induces an action of $G$ on a set of $n-1$ objects, $X$ say.

By the Orbit-Stabiliser Theorem, given $x \in X$ , $\left| \mathrm{Orb}_G(x)\right| \left| \mathrm{Stab}_G(x) \right| = \left| G \right| = n$ . Orbits partition $X$ , so $\left| \mathrm{Orb}_G(x)\right| \leq n-1$ . Hence $\left| \mathrm{Stab}_G(x)\right| \geq \frac{n}{n-1} > 1$ .

Now $\mathrm{Stab}_G(x)$ is a subgroup of $G$ and is non-trivial from above. So $\forall x \in X, P \leq \mathrm{Stab}_G(x)$ . In other words, each element of $P$ fixes all $x \in X$ , so $P \leq \ker(\phi)$ . But this contradicts that $\phi$ is injective. So $G$ is a bountiful group. $\blacksquare$

It is natural to ask whether we can go further, to find all such groups. By perusing resources such as group order statistics (see Element structure of groups of order 32 for example), we are led to investigate the generalised quaternion groups $Q_{2^k}$ . One can consult Keith Conrad’s excellent Generalised Quaternions article for a definition. In particular, we have a remarkable theorem:

Theorem. If a finite $p$ -group has a unique subgroup of order $p$ , then it is cyclic or generalised quaternion.

Proof. See Keith Conrad’s Generalised Quaternions article, Theorem 4.9.

Thus our final result follows:

Classification Theorem for Bountiful Groups: Let $G$ be a Bountiful Group. Then $G$ is isomorphic to one of:
$C_2 \times C_2$
$C_{p^k}$ for some prime power $p^k$ , or
the generalised quaternion groups $Q_{2^k}$ .

Proof. Follows directly from the two theorems above. $\blacksquare$

Acknowledgements

Many thanks to Professor Imre Leader of Trinity College, Cambridge for his support and advice. Thanks also to two of my good friends Yiannis, for his initial investigations with me, and Weida, with whom a discussion led me to consider this idea.

And to the quaternion group $Q_8$ , for being my canonical example to think about!

Michael Ng

Olympiad Tutoring – Piano – Mathematics

Monthly Archives: December 2021

Classifying the Bountiful Groups

Introduction

Notation and Preliminaries

Classification

Acknowledgements