One other question, what is ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##? I'm pretty sure ##\text{Hom}_{\mathbb{K}}(V, W)## is the set of all ##\mathbb{K}##-linear maps from ##V## to ##W##. And we can make ##V## into a ##\mathbb{K}G## module by defining ##g \cdot v = \rho(g)v## I think?? So...
I think I get it now, I had to look up the definition of GL(V). So, ##\rho(g^{-1})## is an automorphism of ##V## and ##\tau(g)## is an automorphism of ##W## and ##\varphi## is a k linear map from ##V## to ##W##.
Sorry for the dumb question but do the ##\circ##'s in 4) mean function composition or matrix multiplication? As I understand it, ##\tau(g)## and ##\rho(g^{-1})## are matrices in ##GL(W)## and ##GL(V)## resp. ?
Thank you for your time and feedback; and sorry for confusion!
Here are two things I would add to make post #20 clearer:
Lemma: Let ##G## be a group and ##H## a subgroup. If ##[G: H] = n##, then there exists a homomorphism ##f : G \rightarrow S_n## with ##\ker f \le H##.
Proof: Let ##X## be...
We know ##y : V \rightarrow \mathbb{R}## is a linear transformation. By Rank-Nullity theorem we have ##\dim V = rank(y) + null(y)##. We note that ##null(y) = \dim W##. If ##rank(y) = 0##, then ##\dim V = \dim W## and so ##\dim W = n##.
If ##rank(y) \neq 0##, then ##rank(y) = 1##. And so ##\dim...
Thank you for your replies. Yes, I think we can use Rank-Nullity theorem, I will try it.
And yeah, I'm happy to use a more standard notation if that's better; I was trying to mimic the textbook.
I am stuck on finding the dimension of the subspace. Here's what I have so far.
Proof: Let ##W = \lbrace x \in V : [x, y] = 0\rbrace##. We see ##[0, y] = 0##, so ##W## is non empty. Let ##u, v \in W## and ##\alpha, \beta## be scalars. Then ##[\alpha u + \beta v, y] = \alpha [u, y] + \beta [v...
We want to show ##\vert (H^\perp)^\perp \vert = \vert H \vert##. We have ##\vert (H^\perp)^\perp \vert = [\widehat{\widehat{G}} : H^\perp] = \frac{\vert\widehat{\widehat{G}}\vert }{\vert H^\perp\vert} = \frac{\vert G \vert}{[G: H]} = \vert H \vert##