# 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 $$\mathbb F_{q^a}\setminus 0$$.

• No, ${\rm SL}(3,4)$ has no subgroups isomorphic to ${\rm SL}(2,8)$, but it has subgroups of that order (which are isomorphic to ${\rm SL}(3,2) \times C_3$). Commented Oct 27, 2019 at 8:27

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 $$a, then $${\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 $$(q^{2b}-1)/(q^2-1)$$.

• Could you please explain why the centralizer of an element of order $r$ in $SL(b,q^a)$ arises from a Singer cycle? Commented Oct 28, 2019 at 16:49
• I have added some explanation. Commented Oct 28, 2019 at 19:31
• thanks you very much! Commented Oct 28, 2019 at 21:35
• Thank you very much for such nice answer. It seems important in the proof that on the right we have SL(b,q^a). Is the statement false with GL(b,q^a)? Or do you have a proof also for that? Commented Oct 29, 2019 at 15:21
• No the statement is also true for ${\rm GL}$ rather than ${\rm SL}$, because (with the exception of $n=2$, $q=2,3$), ${\rm SL}(n,q)$ is the commutator subgroup of ${\rm GL}(n,2)$ and is perfect, so an embedding of ${\rm GL}$ would automatically imply an embedding of ${\rm SL}$. But the above proof does not work directly for ${\rm GL}$. Commented Oct 29, 2019 at 18:32