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Problem. Prove that $\operatorname{GL}_n(k)$ and $\operatorname{SL}_n(k)$ cannot be isomorphic to $S_m$, $m\geq 4$ if $k$ is a finite field with at least two elements.

I am trying to argue by looking at the order of the two linear groups and prove that they cannot be a factorial. I have computed that $$\operatorname{GL}_n(k)=(q^n-1)...(q^n-q^{n-1})$$ and $$\operatorname{SL}_n(k) = \frac{(q^n-1)...(q^n-q^{n-1})}{q-1}$$ where $q=|k|$ and I am taking a look at the special case when $q$ is a prime and conclude the factorial with $q$ occuring $n-2$ times would be much larger than the order of the general linear group.

Is there a better way I could argue this?

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    $\begingroup$ do you know the centres of those groups? $\endgroup$ Oct 27, 2023 at 6:24
  • $\begingroup$ @MatthewTowers Oh yes! The center for the two linear groups are of the same size as $k-\{0\}$ and $\{\omega\in k: \omega^n=1\}$ respectively and the center for $S_n$ is trivial for $n\geq 3$. I think that solves the problem. $\endgroup$
    – user108580
    Oct 27, 2023 at 6:26
  • $\begingroup$ @MatthewTowers That leaves us with only one cases to consider: Can $\operatorname{GL}_n(\Bbb Z/2\Bbb Z)$ be isomorphic to $S_m$ for $m\geq 4$? $\endgroup$
    – user108580
    Oct 27, 2023 at 6:30

1 Answer 1

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Every finite field has at least two elements so that last condition is unnecessary. We can argue as follows. $GL_n(k)$ has center $k^{\times}$ whereas the center of $S_m, m \ge 4$ is trivial, so an isomorphism in the first case can only occur if $k = \mathbb{F}_2$ (and we have $GL_2(\mathbb{F}_2) \cong S_3$ which is why the $m \ge 4$ condition is necessary).

Next, the abelianization of $GL_n(k)$ is known to be $k^{\times}$ (with abelianization map given by the determinant) except when $n = 2$ and $k = \mathbb{F}_2$, so the abelianization of $GL_n(\mathbb{F}_2) \cong SL_n(\mathbb{F}_2)$ for $n \ge 3$ is trivial, but the abelianization of $S_m, m \ge 4$ is $C_2$. So $GL_n(\mathbb{F}_2), n \ge 3$ is never isomorphic to a symmetric group.

The case of $SL_n(\mathbb{F}_q)$ is trickier although the above argument rules out the case $q = 2$. The center of $SL_n(k)$ is the group $\mu_n(k)$ of $n^{th}$ roots of unity in $k$, so for $k = \mathbb{F}_q$ the condition that this group is trivial means that $\gcd(n, q - 1) = 1$. When this happens we have $SL_n(\mathbb{F}_q) \cong PSL_n(\mathbb{F}_q)$ and now we can appeal to the fact that these groups are simple whereas $S_m, m \ge 4$ is not (since it has nontrivial abelianization), unless $n = 2$ and $q = 2, 3$. We ruled out $q = 2$, and when $q = 3$ we have nontrivial center.

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