# Prove that $\operatorname{GL}_n(k)$ and $\operatorname{SL}_n(k)$ cannot be symmetric group?

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?

• do you know the centres of those groups? Oct 27, 2023 at 6:24
• @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. Oct 27, 2023 at 6:26
• @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 $m\geq 4$? Oct 27, 2023 at 6:30

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.