TITLE: Embedding $\mathrm{PGL}(n,q^h)$ in $\mathrm{PGL}(nh,q)$
QUESTION [9 upvotes]: It is not very hard to see that for each prime power $q$ and natural numbers $n,h$, we have an embedding
$$\iota \colon \mathrm{GL}(n,q^h) \hookrightarrow \mathrm{GL}(nh, q),$$
obtained by choosing a basis for the finite field $\mathbb{F}_{q^h}$ over the base field $\mathbb{F}_q$.
It is not very hard either to see that this does not induce an embedding of the corresponding projective groups $\mathrm{PGL}(n,q^h)$ into $\mathrm{PGL}(nh, q)$.
Of course, this does not exclude the possibility that $\mathrm{PGL}(n,q^h)$ is nevertheless isomorphic to a subgroup of $\mathrm{PGL}(nh, q)$, and that is precisely my question:

For which values of $q,n,h$ with $h > 1$ is $\mathrm{PGL}(n,q^h)$ isomorphic to a subgroup of $\mathrm{PGL}(nh, q)$?

Notice that comparing the order of both groups does not rule out anything at all.
I believe I read somewhere that $\mathrm{PGL}(n,q^h)$ is isomorphic to a subgroup of $\mathrm{PGL}(n^h, q)$ (but I forgot the precise construction), so this would at least tell that the answer is positive for $n=h=2$. (Edit: This fact is now confirmed by Derek Holt's answer.)

REPLY [5 votes]: Assume that $n$ is prime to $q^h-1$. Then $\mathrm{SL}(n, q^h)\cap \mathbb F_{q^h}^{\times}$ is trivial and the map $\mathrm{SL}(n, q^h)\rightarrow \mathrm {PGL}(n, q^h)$ is an isomorphism, as $x\mapsto x^n$ induces an iso on $\mathbb{F}_{q^h}^{\times}$. The composition 
$$ \mathrm{SL}(n, q^h)\rightarrow \mathrm{GL}(n, q^h)\rightarrow \mathrm{GL}(nh, q)\rightarrow \mathrm{PGL}(nh, q),$$
where the middle arrow is the one described in the question, is injective, because
the kernel is equal to $\mathrm{SL}(n, q^h)\cap \mathbb F^{\times}_{q}$.