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

Jyrki Lahtonen
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Reyx_0
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    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 ${\rm SL}(3,2) \times C_3$). – Derek Holt Oct 27 '19 at 08:27

1 Answers1

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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 $(q^{2b}-1)/(q^2-1)$.

Jyrki Lahtonen
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Derek Holt
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  • Could you please explain why the centralizer of an element of order $r$ in $SL(b,q^a)$ arises from a Singer cycle? – Ferra Oct 28 '19 at 16:49
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    I have added some explanation. – Derek Holt Oct 28 '19 at 19:31
  • thanks you very much! – Ferra Oct 28 '19 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? – Reyx_0 Oct 29 '19 at 15:21
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    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 ${\rm GL}$. – Derek Holt Oct 29 '19 at 18:32
  • Yes the corrections look good, thanks! – Derek Holt Oct 29 '19 at 21:33