# How is the “Representation of Cosets theorem” a generalization of Cayley's theorem?

Theorem 2.88 (Representation on Cosets). Let $$G$$ be a group, and let $$H$$ be a subgroup of $$G$$ having finite index $$n$$. Then there exists a homomorphism $$\varphi: G \rightarrow S_n$$ with $$\ker \varphi \leq H$$.

The author claims this is a more interesting generalization of Cayley's theorem. However this doesn't seem to really tell anything? If $$\ker \varphi \leq H$$ it could be the case that ker $$\varphi = \{1\}$$, which is Cayley's theorem, so isn't this statement weaker than Cayley's theorem?

• Try \ker\varphi for $\ker\varphi$ – gen-ℤ ready to perish Jun 15 '20 at 1:11
• An often useful consequence of this result is that $\ker\varphi$ is of index at most $n!$ in $G$. And that the kernel is always a normal subgroup. For example, if $[G:H]=3$, then this result implies the existence of a normal subgroup $N\unlhd G$ such that $N\le H$, and $[G:N]\le6$. This comes up frequently when working on finite simple groups. Also, the fact that the image is a subgroup of a very concrete group $S_n$ helps also. Cayley's theorem does not guarantee the existence of non-trivial normal subgroups. I think that the bound on the index of $\ker\varphi$ makes this "more interesting" – Jyrki Lahtonen Jun 15 '20 at 3:47

The only reason to restrict this to subgroups of finite index is that the definition of the symmetric group for an infinite set is not a given. For some, the symmetric group on an infinite set, $$S_X$$, just means all bijections $$X\to X$$; others require the bijections to have finite support (that is, $$\mathrm{supp}(\sigma) = \{x\in X\mid\sigma(x)\neq x\}$$ is finite for each bijection $$\sigma$$), so that the elements of $$S_X$$ can still be described as consisting of a product of disjoint cycles, etc. (Also, its most common application is that a subgroup of finite index contains a normal subgroup of finite index, so the more general statement does not provide wider applications).

If you simply define $$S_X$$ to be the group of bijections $$X\to X$$, a group under composition, then this theorem holds:

Theorem. Let $$G$$ be a group and let $$H$$ be a subgroup of $$G$$. Then there exists a homomorphism $$\varphi\colon G\to S_{G/H}$$ (where $$G/H$$ is the set of left cosets of $$H$$ in $$G$$) with $$\ker(\varphi)\leq H$$.

The special case of this theorem which yields Cayley’s Theorem is $$H=\{e\}$$.

What you've observed is that Cayley's theorem is a special case of this theorem: if $$G$$ is finite then we may pick $$H =\{1\}$$, which forces $$\ker \varphi = \{1\}$$, giving the conclusion of Cayley's theorem. But of course the theorem is more general than this; $$G$$ need not be finite and $$\ker \varphi$$ need not be trivial. That's what Rotman means by "more interesting generalization".

Saying that this is weaker than Cayley's theorem would mean that Cayley's theorem implies this result, which is not the case (of course there's no way to make this precise because both theorems are true).

• The author's statement of Cayley's theorem is for arbitrary groups, not just finite ones. – user736690 Jun 15 '20 at 0:15
• Ah, then probably he means that this is a generalization of Cayley's theorem but specialized to finite groups. Of course it's impossible to use this theorem to prove that an arbitrary group embeds in some symmetric group. – diracdeltafunk Jun 15 '20 at 0:17
• It is more general because it applies in a wider variety of cases (the hypotheses are easier to satisfy). Notably, this isn't anything I can make precise! It's just a phrase we use to convey a general feeling, like how saying that one theorem is stronger than another isn't (usually) saying anything precise: any two theorems are logically equivalent because we have proofs of both! – diracdeltafunk Jun 15 '20 at 0:24
• i realize now that in this theorem in $S_n$ $n$ is the index of subgroup while in Cayley's theorem the isomorphism is into $S_G$, that was where my misunderstanding came from. – user736690 Jun 15 '20 at 0:34
• It is "more interesting" because the conclusion is more general: sometimes the conclusion is exactly the same as in Cayley's theorem, which lets us derive Cayley's theorem as a special case, but sometimes you can conclude something different. For example, this theorem allows you to prove that every finite-index subgroup contains a finite-index normal subgroup -- see @JyrkiLahtonen's comment on the original post. Does that help? – diracdeltafunk Jun 16 '20 at 20:44

In general, the kernel of a $$G$$-action on a set $$X$$ (i.e. the kernel of the equivalent homomorphism $$\lambda:G\to\operatorname{Sym}(X)$$) is given by $$\operatorname{ker}\lambda=\bigcap_{x\in X}\operatorname{Stab}(x)$$.

In particular, if we take the $$G$$-action by left multiplication on the set $$X:=\{gH,g\in G\}$$, then there is a homomorphism $$\lambda:G\to \operatorname{Sym}(G/H)\cong S_{[G:H]}$$, with $$\operatorname{ker}\lambda=\bigcap_{g\in G}gHg^{-1}\le H$$ (see e.g. here).

This result, along with the First Homomorphism Theorem, shows that, among all the subgroups $$H$$ of a given index in $$G$$, an eventual normal one provides the "best focus" (literally, i.e. the smallest image) of $$G$$ in $$S_{[G:H]}$$. In fact, in this case: $$\operatorname{ker}\lambda=H=(\operatorname{ker}\lambda)_{\operatorname{max}}$$.

To recap:

Let's define $$\mathcal{H}_k:=\{H\le G\mid [G:H]=k\}$$. Then, $$\forall H\in\mathcal{H}_k$$, the group $$Q_G(H):=G/\bigcap_{g\in G}gHg^{-1}$$ embeds into $$S_k$$. Moreover, if $$\exists \tilde H\in \mathcal{H}_k$$ such that $$\tilde H\unlhd G$$, then $$|\operatorname{im}(Q_G(\tilde H))|=|\operatorname{im}(G/\tilde H)|=\operatorname{min}\{|\operatorname{im}(Q_G(H))|, H \in \mathcal{H}_k\}$$.