How to prove that $\mathrm{SL}_{2} (\mathbb F_{3})$ is not isomorphic to $S_{4}$? They are both group of order $24$ and both groups have elements of order $2, 3$ and $4$. I don't know how to work from here.
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2This is Exercise 11 in Dummit & Foote's "Abstract Algebra", 3rd, ed. – copper.hat Mar 02 '21 at 07:41
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$SL_2(\mathbb{F}_3)$ has two elements $I, -I$ in its center, but the center of $S_4$ is trivial.
arkeet
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$$A:=\begin{pmatrix}2&0\\1&2\end{pmatrix}\implies A^2=\begin{pmatrix}1&0\\1&1\end{pmatrix}\;,\;\;A^3=\begin{pmatrix}2&0\\0&2\end{pmatrix}=-I\implies A^6=I$$
and thus SL$_2(\Bbb F_3)\;$ has an element of order six....but $\;S_4\;$ hasn't any.
DonAntonio
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@Rikka Experience, using the fact that these matrices are easy to play with, and knowing that this matrix group has a lot of elements of order six (I think $;8;$ such elements...) – DonAntonio Oct 04 '16 at 19:27
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@DonAntonio Please give the presentation first, then it is easy to understand your answer. – MANI Oct 20 '19 at 10:53
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1@MANI Another way to find a matrix of order $6$: the $6$th cyclotomic polynomial is $x^2 - x + 1$, and its companion matrix $\begin{pmatrix} 0 & -1\1 & 1\end{pmatrix}$ has order $6$. – Viktor Vaughn Oct 10 '21 at 03:57
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One can study the subgroups in both cases. The proper subgroups of $SL_2(\mathbb{F}_3)$ are the cyclic groups $C_2,C_3,C_4,C_6$ and the quaternion group $Q_8$. But $S_4$ has no subgroup isomorphic to the quaternions: suppose a subgroup $H \leq S_4$ exists such that $H \cong Q_8$. Now $Q_8$ contains $6$ elements of order $4$, and $S_4$ contains exactly $6$ elements of order $4$, namely the 4-cycles. Thus $H$ contains all $4$-cycles in $S_4$. But then $H$ also contains all products of two $2$-cycles, so $|H| > 8$, a contradiction. Thus no such $H$ exists, and $S_4$ is not isomorphic to $SL_2(\mathbb{F}_3)$.
Of course, also $C_6$ does not arise as a subgroup of $S_4$.
Dietrich Burde
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@Rikka Because every product of two 2-cycles is the square of a 4-cycle: $$(ab)(cd)=(acbd)^2. $$ – anon Oct 04 '16 at 20:34
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@Rikka Compute the powers of $(abcd)$. Also, for $S_4$ we know all subgroups in detail, see here. – Dietrich Burde Oct 04 '16 at 20:36
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The character tables of $S_4$ and of $\mathrm{SL}_2(\mathbb F_3)$ are different. The first group contains only integer entries (as any symmetric group) and the character table of $\mathrm{SL}_2(\mathbb F_3)$ contains a cubic root of unity.
user978144
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