monomorphism from $\mathrm{GL}(n,q^2)$ to $\mathrm{GL}(2n,q)$ Construct an explicit monomorphism from $GL(n,q^2)$ to $GL(2n,q)$. Will $GL(n,q^m)$ always have a subgroup isomorphic with $GL(mn,q^2)$ for every $m,n\in \mathbb{N}$, and prime power $q$?
 A: $\mathbb{F}_{q^m}$ is an $m/n$ dimensional vector space over $\mathbb{F}_{q^n}$ for $n|m$.  multiplication by an element $\alpha\in\mathbb{F}_{q^m}$ gives an $\mathbb{F}_{q^n}$-linear map $\alpha:\mathbb{F}_{q^m}\to\mathbb{F}_{q^m}$, which can be represented by an $m/n\times m/n$ matrix (given a choice of basis for $\mathbb{F}_{q^m}$ over $\mathbb{F}_{q^n}$).  so in terms of matrices, replace each entry $a_{ij}$ in a matrix $A\in{\rm GL}_k(\mathbb{F}_{q^m})$ with an $m/n\times m/n$ matrix from ${\rm GL}_{m/n}(\mathbb{F}_{q^n})$ representing multiplication by $a_{ij}$.  this gives an injection from ${\rm GL}_k(\mathbb{F}_{q^m})$ to ${\rm GL}_{km/n}(\mathbb{F}_{q^n})$.  (replace general linear groups with matrix groups etc if youre nitpicky).
as per another answer consider the $2\times2$ complex matrix
$$
\left(
\begin{array}{cc}
1+i&2+3i\\
-i&3\\
\end{array}
\right)
$$
this becomes a $4\times4$ real matrix 
$$
\left(
\begin{array}{cccc}
1&-1&2&-3\\
1&1&3&2\\
0&1&3&0\\
-1&0&0&3\\
\end{array}
\right)
$$
under the identification of multiplication by $a+bi$ wrt the basis $1,i$ of $\mathbb{C}$ over $\mathbb{R}$ with the matrix
$$
\left(
\begin{array}{cc}
a&-b\\
b&a\\
\end{array}
\right)
$$
A: Can you see how $\mathrm{GL}_n(\mathbb{C})$ is naturally a subgroup of $\mathrm{GL}_{2n}(\mathbb{R})$? You may apply the same idea to the finite fields. 
Hint: A $\mathbb{C}$-linear operator is a priori $\mathbb{R}$ linear.  
I wonder if there is some typo in the second part of the question. As it stands now, it doesn't make much sense, since $\mathrm{GL}_{mn}(\mathbb{F}_{q^2})$ has larger cardinality than $\mathrm{GL}_{n}(\mathbb{F}_{q^m})$.
