3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ This is Exercise 11 in Dummit & Foote's "Abstract Algebra", 3rd, ed. $\endgroup$ – copper.hat Mar 2 at 7:41
10
$\begingroup$

$SL_2(\mathbb{F}_3)$ has two elements $I, -I$ in its center, but the center of $S_4$ is trivial.

$\endgroup$
1
  • 1
    $\begingroup$ Darn, I was gonna say that! $\endgroup$ – D_S Oct 5 '16 at 1:26
6
$\begingroup$

$$A:=\begin{pmatrix}2&0\\1&2\end{pmatrix}\implies A^2=\begin{pmatrix}1&0\\1&1\end{pmatrix}\;,\;\;A^3=\begin{pmatrix}2&0\\0&2\end{pmatrix}=-I\implies A^6=I$$

and thus SL$_2(\Bbb F_3)\;$ has an element of order six....but $\;S_4\;$ hasn't any.

$\endgroup$
3
  • 1
    $\begingroup$ How did you just somehow find this element? $\endgroup$ – Rikka Oct 4 '16 at 19:25
  • $\begingroup$ @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...) $\endgroup$ – DonAntonio Oct 4 '16 at 19:27
  • $\begingroup$ @DonAntonio Please give the presentation first, then it is easy to understand your answer. $\endgroup$ – MANI Oct 20 '19 at 10:53
2
$\begingroup$

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 $S_4$.

$\endgroup$
3
  • $\begingroup$ why must H also contain all products of 2 cycles? $\endgroup$ – Rikka Oct 4 '16 at 20:31
  • $\begingroup$ @Rikka Because every product of two 2-cycles is the square of a 4-cycle: $$(ab)(cd)=(acbd)^2. $$ $\endgroup$ – arctic tern Oct 4 '16 at 20:34
  • $\begingroup$ @Rikka Compute the powers of $(abcd)$. Also, for $S_4$ we know all subgroups in detail, see here. $\endgroup$ – Dietrich Burde Oct 4 '16 at 20:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.