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Can $\operatorname{GL}_n(\Bbb F_{q})$ be isomorphic to $\operatorname{SL}_m(\Bbb F_{r})$, where $\Bbb F_q$, $\Bbb F_r$ are finite fields with $q,r$ elements respectively?

By considering the center, $$\left|\operatorname{Z}(\operatorname{GL}_n(\Bbb F_q))\right|=q-1$$ and $$\left|\operatorname{Z}(\operatorname{SL}_m(\Bbb F_r))\right|=\operatorname{gcd}(m, r-1)$$ and this helps ruling out some cases, but did not solve the problem completely.

Also note that $\operatorname{SL}_m(\Bbb F_r)$ is simple when $\operatorname{gcd}(m,r-1)$ so in this case it could never to isomorphic to the general linear group, which has $\operatorname{SL}_n(\Bbb F_q)$ as a normal subgroup.

As commented below, I listed the order of the two linear groups below:

$$\left|\operatorname{GL}_n(\Bbb F_q)\right|=(q^n-1)\dots (q^n-q^{n-1})$$ and $$\left|\operatorname{SL}_m(\Bbb F_r)\right| = \frac{(r^m-1)\dots (r^m-r^{m-1})}{r-1}$$ I cannot see immediately that the two could not be equal though.

I am guessing they could never be isomorphic except for a finite number of small orders, how could I show it?

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    $\begingroup$ You can compare the orders of the groups itself, then you are done. $\endgroup$ Commented Oct 27, 2023 at 17:00

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The special linear groups are perfect with only two exceptions $\mathrm{SL}_2(2) \cong S_3$ and $\mathrm{SL}_2(3) \cong 2 \cdot A_4$. On the other hand if $\mathrm{GL}_n(q)$ is perfect then $q = 2$ and $\mathrm{GL}_n(2) = \mathrm{SL}_n(2)$.

It is more interesting to ask about coincidences between PSL's. The only coincidences turn out to be $\mathrm{PSL}_2(4) \cong \mathrm{PSL}_2(5) \cong A_5$ and $\mathrm{PSL}_2(7) \cong \mathrm{PSL}_3(2)$. Otherwise no two PSL's have the same order, and to prove this one usually uses a theorem of Zsigmondy.

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  • $\begingroup$ May I ask about the coincidences between SL's? I can conclude $\operatorname{SL}_m(p)\simeq \operatorname{SL}_n(q)$ implies $(m,p)=(n,q)$. Can I show different SL's are never isomorphic to each other? $\endgroup$
    – user108580
    Commented Oct 27, 2023 at 18:03
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    $\begingroup$ @user108580 ${\rm SL}_m(p) \cong {\rm SL}_n(q)$ would imply ${\rm PSL}_m(p) \cong {\rm PSL}_n(q)$ by considering $G/Z(G)$, so in fact there are no such isomorphisms. $\endgroup$
    – Derek Holt
    Commented Oct 27, 2023 at 18:43
  • $\begingroup$ @DerekHolt Oh yes. I wasn't thinking about projective linear groups. Thank you! $\endgroup$
    – user108580
    Commented Oct 27, 2023 at 18:46

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