## Lifting and Descending Representations

Let’s recall that a group representation is, among other things, a group homomorphism. This has a few consequences.

First of all, we can consider the kernel of a matrix representation . This is not the kernel we’ve talked about recently, which is the kernel of a -morphism. This is the kernel of a group homomorphism. In this context, it’s the collection of group elements so that the image is the identity transformation. We call this subgroup . If is the trivial subgroup, we say that is a “faithful” representation, since it doesn’t send any two group elements to the same matrix.

Now, basic group theory tells us that is a normal subgroup, and so we can form the quotient group . I say that the representation “descends” to a representation of this quotient group. That is, we can define a representation by for all cosets . We have to check that this doesn’t depend on which representative of the coset we choose, but any other one looks like for . Then , since is the identity matrix by definition.

I say that is a faithful representation. That is, the only coset that sends to the identity matrix is the one containing the identity: itself. And indeed, if , then , and so and in the first place.

Next, is irreducible if and only if is. We’ll prove this one by using the properties we proved a few days ago. In particular, we’ll calculate the inner product of the character of with itself. Writing for the character of and for that of , we find that

and we use this to calculate:

Essentially, the idea is that each group element in the kernel contributes the same to the sum, and this is exactly compensated for by the difference between the sizes of and . Since the two inner products are the same, either both are or neither one is, and so either both representations are irreducible or neither one is.

We can also run this process in reverse: let be a finite group, let be any normal subgroup so we have the quotient group , and let be any representation of . We will use this to define a representation of the original group by “lifting” the representation .

So, the obvious choice is to define . This time there’s no question that is well-defined, since here we start with a group element and find its coset, rather than starting with a coset and picking a representative element. And indeed, this is easily verified to be a representation.

If is faithful, then the kernel of is exactly . Indeed, if , then . If is faithful, then we must have , and thus .

And finally, is irreducible if and only if is. The proof runs exactly as it did before.