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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...

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  • $\begingroup$ According to GroupProps, you can prove it with Frobenius reciprocity. $\endgroup$ – runway44 Jun 17 at 2:54
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"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 $\rho (x)\rho (x^{-1})=\rho(e)=I$.

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