# Abstract Nonsense

## Cayley’s Theorem

Point of post: In this post we give a brief discussion of the permutation group on an arbitrary set and prove Cayley’s theorem. We will then show it’s use by proving a fact about complex characters (something that will come up much when we talk about representation theory, soon enough
!)

Motivation

Group theory arose from studying permutation groups (the set of all bijections on a set with function composition as the binary operation). Cayley’s theorem will tell us that this wasn’t a bad start. Indeed, Cayley’s theorem will show that every group is isomorphic to a subgroup of the permutation group on the underlying set. This, in a nutshell, is huge. This is one of the most powerful properties of groups. This will become clear when we apply Cayley’s theorem to the basic theory of complex characters.

Permutation Group

I have already defined this elsewhere, but for didactic purposes I think it is well-suited to repeat this.

Let $\Omega$ be a set, and define the permutation group on $\Omega$, denoted $S_{\Omega}$ or $\text{Perm}(\Omega)$, to be the set

$S_{\Omega}=\left\{\pi:\Omega\to\Omega\mid\pi\text{ is a bijection}\right\}$

with function composition. It’s easy to check that $S_{\Omega}$ is, in fact, a group. This leads us to Cayley’s theorem

Theorem(Cayley): Let $G$ be a group and for every $g_0\in G$ define the map $\theta_{g_0}:G\to G:g\mapsto g_0g$. Then,

\displaystyle \begin{aligned}&\mathbf{(1)}\quad \theta_{g_0}\in S_G\\ &\mathbf{(2)}\quad \Theta:G\to S_G:g\mapsto\theta_g\textit{ is an embedding}\end{aligned}

Proof:

$\mathbf{(1)}$: To see that $\theta_{g_0}$ is injective we merely note that if $\theta_{g_0}(g)=\theta_{g_0}(h)$ then $g_0g=g_0h$ and thus $g=h$. Furthermore, to see that $\theta_{g_0}$ is surjective we note that if $g\in G$ then $\theta_{g_0}\left(g_0^{-1}g\right)=g_0g_0^{-1}g=g$ from where surjectivity, and thus bijectivity, follows.

$\mathbf{(2)}$: We first prove that $\Theta$ is a homomorphism. To do this we note that for any fixed $g_1,g_2\in G$ and variable $g\in G$ we have that

$\left(\Theta(g_1)\circ\Theta(g_2)\right)(g)=\left(\Theta(g_1)\right)\left(\Theta(g_2)\right)(g)=\left(\Theta(g_1)\right)(g_2g)=g_1g_2g=\left(\Theta(g_1g_2)\right)(g)$

and since $g$ was arbitrary it follows that $\Theta(g_1)\Theta(g_2)=\Theta(g_1g_2)$. Now, to prove that $\Theta$ is injective we merely note that if $g\in\ker\Theta$ then $\theta_g=\text{id}_G$ so that $\theta_g(e)=g=e$. Thus, $\ker\Theta=\{e\}$. Thus, we are done. $\blacksquare$

Corollary: Let $G$ be a group and $G'$ an abelian group. Then, if $\varphi\in\text{Hom}\left(G,G'\right)$, then

$\displaystyle \sum_{g\in G}\varphi(g)=\sum_{g\in G}\varphi(gg_0)$

for every $g_0\in G$.

Proof: By Cayley’s theorem we have that $\theta_{g_0}$ is a bijection and so each term of $\displaystyle \sum_{g\in G}\varphi(g)$ appears precisely once in $\displaystyle \sum_{g\in G}\varphi(gg_0)$ from where the conclusion follows. $\blacksquare$

So from this we can prove an interesting fact about complex characters on finite groups, but first we must define these concepts. Let $G$ be an abelian group. Then, a complex character on $G$ is a homomorphism $\chi:G\to\mathbb{C}^{\times}$ where $\mathbb{C}^{\times}$ is the set of non-zero complex numbers under multiplication.

Theorem: Let $G$ be a finite abelian group and $\chi,\psi$ two complex characters on $G$. Then,

$\displaystyle \sum_{g\in G}\chi(g)\psi\left(g^{-1}\right)=\begin{cases}|G| & \mbox{if}\quad \psi=\chi\\ 0 & \mbox{if}\quad \psi\ne\chi\end{cases}$

Proof: If $\chi=\psi$ this is clear since our sum reduces to $\displaystyle \sum_{g\in G}1$. Suppose then that $\chi\ne\psi$. Then, there exists some $g_0\in G$ such that $\chi(g_0)\ne \psi(g_0)$ or, said differently, $\chi(g_0)\psi\left(g_0^{-1}\right)\ne 1$. Note then that

\displaystyle \begin{aligned}\chi(g_0)\psi\left(g_0^{-1}\right)\sum_{g\in G}\chi(g)\psi\left(g^{-1}\right) &= \sum_{g\in G}\chi\left(gg_0\right)\psi\left((gg_0)^{-1}\right)\\ &= \sum_{g\in G}\chi(g)\psi\left(g^{-1}\right)\end{aligned}

(where we’ve made use of our previous corollary in the last step). Thus, upon subtraction we get

$\displaystyle \left(\chi(g_0)\psi\left(g_0^{-1}\right)-1\right)\sum_{g\in G}\chi(g)\psi\left(g^{-1}\right)=0$

and since $\chi(g_0)\psi\left(g_0^{-1}\right)-1\ne 0$ we may conclude that

$\displaystyle \sum_{g\in G}\chi(g)\psi\left(g^{-1}\right)=0$

as desired. $\blacksquare$

Corollary: Let $G$ be a finite group and $\chi$ be a complex character on $G$. Then,

$\displaystyle \sum_{g\in G}\chi(g)=\begin{cases}|G| & \mbox{if}\quad \chi=1\\ 0 & \mbox{if}\quad \chi\ne 1\end{cases}$

where $1$ is the trivial homomorphism $1:G\to \mathbb{C}^{\times}:g\mapsto 1$.

Proof: Take $\psi=1$ in the previous theorem.

References:

1.  Lang, Serge. Undergraduate Algebra. 3rd. ed. Springer, 2010. Print.

2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print

December 28, 2010 -

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6. Hi my friend. I very much like your blog and the idea behind it. I hope to read much of it and be inspired by it ! I’m pretty much (I guess !) on the same math level as you, this is my 4th year of university, getting in some interesting cohomology of groups, Higher K-Theory and Riemmanian Geometry.
Anyway ! I think there are two typos in your post : In the statement of Cayley’s Theorem, I think $\Theta$ is only an embedding and not an iso. Also, in the proof, at the end you say : let’s show it is surjective, and you show the kernel is 0 ^^
See ya !

Comment by zefiloux | November 14, 2011 | Reply

• Hello friend! It’s good to hear that my little blog is of any interest of you. I you are doing K-theory I wouldn’t say we’re quite on the same level, unless what you meant is that by the time I’m a senior (next year) we’ll be on the same level (which is my goal). Right now I’m just doing some homological algebra/group cohomology and some of the abelian category stuff that goes along with it. Jealous of the Riemannian Geometry stuff! Always wanted to learn a little–I’m taking a basic, undergraduate diff geo course right now, it’s not very fulfilling.

Thanks for pointing out those mistakes! I really rushed through these basic group theory posts since they weren’t of that much interest of me, so they are doubtlessly riddled with typos and logical errors.

I would love to learn more about your background/interests (I love talking to other math enthusiasts). You should shoot me an e-mail at alex.youcis@gmail.com

Regardless, thanks again!

Best,
Alex

Comment by Alex Youcis | November 14, 2011 | Reply

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