Let $T: G \to GL(V)$ be a representation of some group $G$ on a vector space $V$ with a $G$-invariant subspace $W$. In part of a proof I am doing, I am showing that if $T(g)(v-u) \in W$, then $v-u \in W$.
I have argued as follows:
It may not be obvious why $T(g)(v - u) \in W \implies v-u \in W$. Recall that $W$ is $G$-invariant. Then if we consider the action of $T(g^{-1})$ on $T(g)(v - u) \in W$ we note that we must have the output in $W$ (since $W$ is, in particular, invariant under $T(g^{-1}))$. This output is $T(g^{-1})T(g)(v - u) = v-u$.
However, I have a (sinking) feeling that I should be able to do this more cleanly. In particular, I feel like I shouldn't need to make reference to $T(g^{-1})$, and should instead be able to directly use that $T(g)$ is an automorphism of $W$ to prove this. Can someone point out what I am missing? It feels like I should just be able to use properties of bijections on a subdomain to argue this.
If you already know a priori that $T(g)$ defines an automorphism of $W$, then you can use surjectivity and pick $w \in W$ such that $T(g)(w) = T(g)(v - u)$. Then $w = v - u$ since $T(g)$ is injective.
P.S. How do you know that $T(g)$ defines an automorphism of $W$? I'm guessing it's because you know that $T(g^{-1}) = T(g)^{-1}$. At some point you have to use the fact that $T$ is a homomorphism.