sub-$G$-representations "So let $G$ be a finite group, $H$ a proper, nontrivial normal subgroup of $G$. For any representation $\rho: G \to \text{GL}(V)$ define the $H$-invariants of $V$ as $$V^H := \{v \in V \text{ }|\text{ } \rho(h)(v) = v \text{ for all }h \in H\}.$$Show that $V^H$ is a sub-$G$-representation of $V$."
Here is a start at a proof, I am not sure if it is right or wrong, help appreciated.
So we show if $V^H$ is an invariant subspace of $V$, then $gV^H = V^H$ for all $g \in G$. Because group elements are invertible, their operations on $V$ are invertible, hence $gV^H$ and $V^H$ have the same dimension. If $gV^H \subset V^H$, then $gV^H = V^H$. Is that all there is to it?
Also, I am curious as to whether such a sub-representation has to be trivial or not.
 A: 
If $gV^H\subseteq V^H$ then $gV^H=V^H$. Is that all there is to it?

There are three issues with this argument. First, your implication doesn't follow: $A\subseteq B$ and the fact that the subspaces $A$ and $B$ have the same dimension does not imply $A=B$ if they are infinite-dimensional spaces. Second, you never actually showed your hypothesis $gV^H\subseteq V^H$ (for all $g\in G$) is true. And then thirdly, you never used the hypothesis that $H$ is a normal subgroup, which is necessary.
It suffices to show that $gV^H\subseteq V^H$ for all $g$, since if both $g$ and $g^{-1}$ map $V^H$ into itself, they must be mutually inverse, hence $g$ (and $g^{-1}$) are invertible maps on $V^H$, so that $gV^H=V^H$. How are you going to show the inclusion $gV^H\subseteq V^H$ though? Unpackage its meaning. If $v\in V$ is $H$-invariant, then you must show (for an arbitrary $g\in G$) that $gv$ is also $H$-invariant. Here is where the normality of $H$ comes into play. Can you see how?
As for whether $V^H$ must be trivial, of course not. It should be pretty doable to pick any group $G$ with a normal subgroup $H$ and then construct a representation $V$ of $G$ for which $V^H$ is nontrivial. Indeed, have you heard of the regular representation? Not to mention we could arrange for $V^H$ to be a direct sum of trivials, so that it's technically nontrivial. (To do this easily, make $V$ itself a sum of trivials.)
A: Let $v\in V^{H}$. As $H \triangleleft G$, we have $g^{-1} h g \in H$ for any $g\in G$ and $h\in H$. Consequently, $$\rho(g^{-1} h g)(v) = v \implies \rho(h) \cdot \rho(g) (v) = \rho(g)(v).$$ Therefore $H$ fixes $\rho(g)(v)$, hence $\rho(g)(v) \in V^H$. It follows $V^H$ is $G$-invariant.
The sub-representation $V^{H}$ need not be trivial. Let $G$ be a group and $H\triangleleft G$. Consider the regular representation $V = \langle \{e_g \, \vert \, g\in G\} \rangle$. If $h_1$, $h_2$, $\dots$, $h_k$ are the elements of $H$, note that for any $h\in H$, $$\rho(h)(e_{h_1} + \dots + e_{h_k} ) = e_{h\cdot h_1} + \dots + e_{h\cdot h_k} = e_{h_1} + \dots + e_{h_k}.$$ Thus, $e_{h_1} + \dots + e_{h_k} \in V^{H}$, implying $V^{H}$ is not the zero subspace of $V$ $($as $e_{h_1} + \dots + e_{h_k}$ is nonzero$)$. 
To prove $V^{H}$ is not trivial, we will need to show that $G$ does not fix $V^{H}$. For any $g\in G$ with $g\not\in H$, we note $$\rho(g)(e_{h_1} + \dots + e_{h_k}) = e_{g\cdot h_1} + \dots + e_{g \cdot h_k},$$ which cannot equal $e_{h_1} + \dots + e_{h_k}$. For if $$e_{g\cdot h_1} + \dots + e_{g \cdot h_k}  = e_{h_1} + \dots + e_{h_k},$$ linear independence of the basis $\{e_g\}$ would tell us each $e_{g\cdot h_i} = e_{h_j}$ for some $j$. But then $g\cdot h_i = h_j$, which means $g\in H$, contrary to the choice of $g$.
