What does it mean for a subgroup to fix a vector space? I was wondering if anyone could explain what is meant by the two statements below? Any help would be greatly appreciated.
Let $U=\mathbb{F}_{q}^{m}$ and $W=\mathbb{F}_{q}^{n}$ be two vector spaces of dimension $m$ and $n$ over $\mathbb{F}_{q}$, respectively. Let $V_{1}=U \oplus W=\mathbb{F}_{q}^{m+n}$ and $V_{2}=U \otimes V=\mathbb{F}_{q}^{mn}$. Obviously, this is very simple but it then says:
Let $G$ be the subgroup of $GL(V_{1})=GL(m+n,\mathbb{F}_{q})$ that fixes $U$. Let $H$ be the group $(\mathbb{F}_{q}^{+})^{mn}:(GL(m,\mathbb{F}_{q})\times GL(n,\mathbb{F}_{q}))$ induced by the natural action of $GL(m,\mathbb{F}_{q})\times GL(n,\mathbb{F}_{q}))$ on the tensor product $V_{2}$. What is meant by these two statements? 
 A: If a group acts on a set $X$ and $Y$ is a subset of $X$ then I would nornally interpret the "subgroup of $G$ that fixes $Y$" to mean $\{ g \in G : g(y) \in Y\,\forall y \in Y \}$. There is also the possibility that it could mean $\{ g \in G : g(y) = y\,\forall y \in Y \}$, but my guess is that this is not what is meant here. In your example, $X$ is the vector space $V_1$ and $Y$ is the subspace $U$.
For the second question, there is a natural induced action, $\phi$ say, of ${\rm GL}(m,q) \times {\rm GL}(n,q)$ on the tensor product $U \otimes V$ defined simply by $g \in {\rm GL}(m,q) \times {\rm GL}(n,q)$ maps $u \otimes v$ to $g(u) \otimes g(v)$, but I am afraid that I cannot make much sense of the description of $H$ as $({\mathbb F}_q^+)^{mn}:({\rm GL}(m,q) \times {\rm GL}(n,q))$. Perhaps this refers to the semidirect product $V_2 \rtimes ({\rm GL}(m,q) \times {\rm GL}(n,q))$ where the action of ${\rm GL}(m,q) \times {\rm GL}(n,q)$ on $V_2$ is given by the image of $g$ in ${\rm GL}(V_1)$ defined by the above action $\phi$. That would make sense!
