# When is general linear group isomorphic to special linear group?

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?

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

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.
• 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? Commented Oct 27, 2023 at 18:03
• @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. Commented Oct 27, 2023 at 18:43