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
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.
I have already defined this elsewhere, but for didactic purposes I think it is well-suited to repeat this.
Let be a set, and define the permutation group on , denoted or , to be the set
with function composition. It’s easy to check that is, in fact, a group. This leads us to Cayley’s theorem
Theorem(Cayley): Let be a group and for every define the map . Then,
: To see that is injective we merely note that if then and thus . Furthermore, to see that is surjective we note that if then from where surjectivity, and thus bijectivity, follows.
: We first prove that is a homomorphism. To do this we note that for any fixed and variable we have that
and since was arbitrary it follows that . Now, to prove that is injective we merely note that if then so that . Thus, . Thus, we are done.
Corollary: Let be a group and an abelian group. Then, if , then
for every .
Proof: By Cayley’s theorem we have that is a bijection and so each term of appears precisely once in from where the conclusion follows.
So from this we can prove an interesting fact about complex characters on finite groups, but first we must define these concepts. Let be an abelian group. Then, a complex character on is a homomorphism where is the set of non-zero complex numbers under multiplication.
Theorem: Let be a finite abelian group and two complex characters on . Then,
Proof: If this is clear since our sum reduces to . Suppose then that . Then, there exists some such that or, said differently, . Note then that
(where we’ve made use of our previous corollary in the last step). Thus, upon subtraction we get
and since we may conclude that
Corollary: Let be a finite group and be a complex character on . Then,
where is the trivial homomorphism .
Proof: Take in the previous theorem.
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