Cayley's Theorem (in my class): Let $G$ be a finite group of order $n$. Then, $G$ is isomorphic to a subgroup of $S_n$.

I trying my best to understand the proof of this theorem, but can't seem to fully grasp it. My biggest problem area is understanding what exactly is described by $\phi$, the isomorphism used in the proof. Can someone describe $\phi $ both informally and formally?

  • $\begingroup$ What is the proof that was presented to you ? $\endgroup$ – Astyx Oct 29 '16 at 21:38
  • $\begingroup$ Quicky: for any $\;g\;$, the $\;n\;$ products $\;gx\;,\;\;x\in\Bbb G\;$ define a permutation on the $\;n\;$ letters that are the elements of $\;G\;$ . You identify the group of permutations with $\;S_n\;$ and the above permutation defined by $\;g\;$ via the homomorphism $\;\phi\;$ . There must be tens of thousands of pages in the web which deal with this, and there must just be one that you will find nice to read and understand. $\endgroup$ – DonAntonio Oct 29 '16 at 21:39

Here's an informal argument:

You want to show that $G$ is isomorphic to a subgroup of $S_n$ (for $n=\vert G\vert$). Well, how do you show that something is isomorphic to a subgroup of $S_n$? Actions! Elements of $S_n$ are understood by how they permute the set $[n]=\{1, 2, . . . , n\}$. So in order to show that $G$ is isomorphic to a subgroup of $S_n$, what we need is a machine for doing the following:

  • Take in a $g\in G$.

  • Spit out a permutation $\pi_g\in S_n$.

The hope is that the map $h: G\rightarrow S_n: g\mapsto \pi_g$ will be a homomorphism from $G$ to $S_n$, and that it will be injective; then $G$ will be isomorphic to $im(h)$.

So this boils down to:

How can I think of an element of $g$ as permuting some $n$-element set?

This is hard because $G$ is an abstract group - we have no sense of what $G$ is "meant to do." So we have to cook up some interpretation of $G$ from scratch!

Luckily, there is a natural $n$-element set lying around: $G$ itself! So, is there a way we can think of an element of $G$ permuting the whole set $G$?

The answer is: yes! Given an element $g$, consider the map from $G$ to $G$ defined as $$\pi_g: a\mapsto g\cdot a.$$ Each $\pi_g$ is a permutation of $G$, we have $\pi_g=\pi_h$ iff $g=h$, and $\pi_g\circ \pi_h=\pi_{gh}$ (do you see why these facts are true?); so the map $h: g\mapsto \pi_g$ is in fact an injective group homomorphism from $G$ to the group of permutations of $G$ . . .

. . . but this latter group is just $S_n$, renamed!

  • 2
    $\begingroup$ I loved the easy flowing writing style, it takes an effort to write this clearly. $\endgroup$ – Sergio Parreiras Oct 30 '16 at 20:55

For $g\in G$, denote $\phi_g:x\mapsto gx$

$\phi_g$ is a permutation of G.

What's more $\phi_g \circ\phi_{g'} = \phi_{gg'}$ and $\phi_e = id$, thus $\{\phi_g, g\in G\}$ is a subgroup of $\mathfrak S(G)$ (the set of permutations of $G$)

Now consider the function $f:g\mapsto\phi_g$. We will prove that $f$ is an isomorphism and this will be Cayley's Theorem.

First see that $f(g)\circ f(g') = f(gg')$ (see above)

Then note $f(g)=id \iff g=e$ thus $\ker f = e$ and therefore $f$ is injective. Since $G$ is finite of order n, and so is $\{\phi_g, g\in G\}$, what follows is that $f$ is bijective and is therefore an isomorphism.

Thus $G$ and $\{\phi_g, g\in G\}$ are isomorphic, and since $\{\phi_g, g\in G\}$ is a subgroup of $\mathfrak S(G)$, we deduce that G is isomorphic to a subgroup of $S_n$


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