# Why is dim(ρW)=dimW for ρ being a linear representation?

first time asking on here so sorry if it's a bit clumsy.

I will have to hold a seminar tomorrow and my professor helped me with one of my proofs. The proof is about showing that for any group $$G$$ and any abelian Subgroup $$A$$ the following statement is true:

The degree of any irreducible representation of $$G$$ is less or equal to $$\frac{g}{a}$$ with $$g$$ being the order of $$G$$ and $$a$$ being the order of $$A$$.

I've already shown that any irreducible subrepresentation of an abelian group has degree $$1$$. Later in my proof my professor wrote something along those lines:

$$\dim(\rho*W) = \dim(W)$$ for $$\rho$$ being a linear representation of $$G$$ in $$V$$ and $$W$$ being a subspace of $$V$$ and $$W$$ being the representation space of $$A$$ (abelian subgroup of $$G$$). He wrote something about $$\rho$$ being an isomorphism but as far as I know, linear representations usually are only (Group-)Homomorphisms but not necessarily Isomorphisms. So can anybody please explain to me why this statement is true in this general case, if it's even true at all?

I would be very thankful if anybody could explain this to me. My main problem is that $$\rho$$ (referring to any element of $$G$$) is more or less a Matrix. Multiplying this matrix with $$W$$ (so every element in $$W$$) must not necessarily produce a vector space with the same dimension as the space $$W$$. For example let's look at the matrix with $$0$$ in each entry. Multiplying this matrix with any vector in $$W$$ would give me only the vector space containing $$0$$ which is most certainly not $$W$$.

I'm sorry to bother you and I hope my problem isn't something completely obvious, I've been missing for hours now...

• According to GroupProps, you can prove it with Frobenius reciprocity. – runway44 Jun 17 at 2:54

"He wrote something about $$\rho$$ being an isomorphism but as far as I know, linear representations usually are only (Group-)Homomorphisms but not necessarily Isomorphisms."
The problem is you're being a little sloppy about exactly what a representation is. If $$\rho$$ is a representation of $$G$$ on $$V$$ there's no such thing as $$\rho W$$.
In fact $$\rho$$ is precisely a homomorphism $$\rho :G\to GL(V)$$, where $$GL(V)$$ is the group of invertible (bounded?) linear maps from $$V$$ to itself. So there's no such thing as $$\rho W$$; instead you could talk about $$\rho (x)W$$ for $$x\in G$$.
Yes, $$\rho$$ is just a homomorphism. But for each $$x\in G$$, $$\rho (x)$$ is an automorphism of $$V$$.
When you say $$\rho$$ is more or less a matrix of course you mean $$\rho (x)$$. If the definition didn't specify invertibility as above but allows any matrix note that $$\rho (x)$$ is automatically invertible, since $$\rho (x)\rho (x^{-1})=\rho(e)=I$$.