The Dual Group of the Product is the Product of the Dual Groups
Point of post: In this post we show that given a finite collection of finite groups the dual group of their product is the product of their dual groups.
Often of course groups come naturally to us as the product of other groups. Of course it would be nice if would tell something about the dual group of the full group if we knew the dual group of each of the product factors. In this post we show, once again, that everything carries over just as we would hope and that the product of the dual groups is completely determined by the dual groups of each of its factors.
The Dual Group of the Product of Groups
We now show that in essence the dual group of the factors of a product of groups entirely determines the dual group for the full product. In fact, we show that the product of dual groups is equal to the dual of the product of the groups. Indeed,
Theorem: Let and be finite groups, then
Proof: Define by where is the tensor product of and . Clearly this map is well-defined in the sense that for every since . To see this is a homomorphism we first note that
That said, note that for every one has that
so that the characters of and are the same, and thus they are the same. So,
Thus, is a homomorphism.
To see that is injective suppose that then is the trivial character. Thus, let be given then and thus so that . Using the exact same techinque shows that so that from where it follows that is trivial and thus is injective.
Finally, to see that is surjective we note by previous theorem that any is of the form for and . But, we have that and since we may conclude that so that and and so and thus and since was arbitrary surjectivity follows. The conclusion follows.
Clearly by induction it follows that:
Corollary: Let be finite groups then,
1. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Math. Soc., 1996. Print.