Why $\dim V^{G} = \dim\operatorname{Hom}_{G}(\mathbb{C}, V)$? Let $G$ be a finite group. Let $V^{G} = \{v\in V: \pi(g)v=v, \forall g\in G\}$ be the subspace of invariants, where $(\pi, V)$ is a representation of $G$. 
Why $\dim V^{G} = \dim \operatorname{Hom}_{G}(\mathbb{C}, V)$? By Schur Lemma, $\dim \operatorname{Hom}_{G}(\mathbb{C}, V)$ is the number of trivial representations of $G$ which occur in $V$. Thank you very much.

A map $f$ in $\hom_G(\mathbb{C},V)$ is a function $f:\mathbb{C}\to V$ which is $G$-linear. We can write this explicitly as $f(g.\lambda)=gf(\lambda)=\lambda g f(1)$. Since $f(g.\lambda)=\lambda f(g.1)$ we can cancel out $\lambda$ to get $f(g.1)=g f(1)$ but the action of $G$ on $\mathbb{C}$ is trivial so that $f(g.1)=f(1)$. We conclude that
$$f(1)=g.f(1)$$
This implies that $f(1)\in V^G$. It is easy to see that this map is injective. We can show that it is surjective: take $v\in V^G$ and define $f(\lambda)=\lambda v$ (this is inspired in the previous argument). You can check easily that this $f$ is $G$-linear and that this construction is the inverse of the previous one.
