Is there an isomorphic copy of $SL(a,q^b)$ in $SL(b,q^a)$ Let $q$ be a prime power and $a,b$ be positive coprime integers.  Let $SL(a,q^b)$ be the special linear group of $a\times a$ matrices over the field $\mathbb F_{q^b}$. 
Is it true that there is no isomorphic copy of $SL(a,q^b)$ in $SL(b,q^a)$?
I can show it by cardinality arguments if $a>b$ but I am not sure on how to do the other case. The general argument seems to be that the map has to be both $\mathbb F_{q^a}$ and $\mathbb F_{q^b}$ linear, so it has to be $\mathbb F_{q^{ab}}$ linear, but since the inclusion is purely abstract you cannot really conclude that.
You can assume that $q$ is large enough if needed and/or that the isomorphic copy has to be transitive of $\mathbb F_{q^a}\setminus 0$. 
 A: When $a=1$, ${\rm SL}(1,q^b)$ is trivial and so it is of course a subgroup of ${\rm SL}(b,q)$.
Otherwise, if $a$ and $b$ are coprime with $a<b$, then ${\rm SL}(a,q^b)$ is not isomorphic to a subgroup of ${\rm SL}(b,q^a)$.
This is easy to show when $a>2$. By Zsigmondy's theorem, there is a prime $r$ that divides $q^{b(a-1)}-1$ that does not divide $q^k-1$ for any $k < b(a-1)$. Then $r$ divides the order of ${\rm SL}(a,q^b)$ but, since $a$ is coprime to $b$ and to $a-1$, we see that $at$ cannot be a multiple of $b(a-1)$ for any $t \le b$, and so $r$ does not divide the order of ${\rm SL}(b,q^a)$.
When $a=2$, the result cannot be deduced immediately from Lagrange's theorem. But ${\rm SL}(2,q^b)$ has elements of order $q^b+1$, which are the intersections of Singer cycles of ${\rm GL}(2,q^b)$ with ${\rm SL}(2,q^b)$. With the exception of the case $q=2$, $b=3$ (which can be dealt with separately by a computer calculation), there is a Zsigmondy prime $r$ dividing $q^{2b}-1$ but not  $q^t-1$ for any $t<2b$, and $r$ divides $q^b+1$.
Now the centralizer of  an element of order  $r$ in ${\rm SL}(b,q^2)$ again arises from a Singer cycle, but it has order $(q^{2b}-1)/(q^2-1)$, which is not divisible by $q^b+1$ (recall that $b$ is odd). So ${\rm SL}(b,q^2)$ has no element of order $q^b+1$, and there is again no embedding.
To understand the centralizer of an element of order $r$, note that the  cyclic subgroup spanned by such an element acts irreducibly on the natural module of ${\rm SL}(b,q^2)$, and so by Schur's lemma, and the fact that finite division rings are fields, its centralizer in ${\rm GL}(b,q^2)$ is isomorphic to the multiplicative group of a finite field of order a power of $q$. Since its order is divisible by $r$, it must be cyclic of order $q^{2b}-1$, so it is a Singer cycle in ${\rm GL}(b,q^2)$. Its intersection with ${\rm SL}(b,q^2)$ has order $(q^{2b}-1)/(q^2-1)$.
