7
$\begingroup$

This is a problem from Ph.D. Qualifying Exams.

Show that the symmetric group $S_6$ contains two subgroups that are isomorphic to $S_5$ but are not conjugate to each other.

Here is my method. $S_5$ contains 6 Sylow 5-subgroup, and $S_5$ act by conjugation on the 6 groups transitively by Sylow's 2nd Theorem, therefore induces a homomorphism $\phi: S_5 \to S_6$. Now, if $ker\phi$ is trvial, then $Im\phi$ is a subgroup of $S_6$ isomorphic to $S_5$. Since $Im\phi$ is a transitive subgroup, it is not conjugate to the subgroup which permutes 5 letters fixing 1 letter, i.e., the subgroup $Sym\{1,2,3,4,5\}\cong S_5$.

So what we need to do is to show $ker\phi$ is trvial. Taken one of 6 Sylow 5-subgroups $H$, there are 6 conjugation orbits, so $N_G(H)$, the normalizer of $H$ has order $20=\frac{120}{6}$ by counting formula. Here I stopped. I know if I can show the intersection of the 6 normalizers contains only identity element, it will be done, but how can I proceed?

$\endgroup$
  • 3
    $\begingroup$ The 5-Sylow subgroups are all conjugate to each other, hence $Im\varphi$ has order at least $6$. That means that $Ker\varphi$ has order no greater than $20$, which means it is trivial. $\endgroup$ – lulu Aug 4 '15 at 2:32
  • $\begingroup$ @lulu Why do you say $Ker\phi$ has order no greater than 20, then it is trivial? $\endgroup$ – Yilong Zhang Aug 4 '15 at 2:41
  • 1
    $\begingroup$ Because $Ker\varphi$ is normal. $S_5$ hasn't got any non-trivial normal subgroups further down than index $2$. $\endgroup$ – lulu Aug 4 '15 at 2:43
  • $\begingroup$ @lulu Got it, thanks! $\endgroup$ – Yilong Zhang Aug 4 '15 at 2:44
  • $\begingroup$ Side note: I was asked this question at my Phd orals, some while back. $\endgroup$ – lulu Aug 4 '15 at 2:45
2
$\begingroup$

As requested, here is a summary of the points raised in the comments.

The argument as posted in the question is very nearly complete.

As is true for any homomorphism, $Ker(\phi)$ is a normal subgroup. Of course, $S_5$ really hasn't got any normal subgroups. There is $S_5$ itself, the trivial group, and $A_5$. Now, as was sorted out in the question itself, the fact that all the $5$-Sylow subgroups are conjugate implies that $Im(\phi)$ is non-trivial, it has to have at least $6$ elements! It follows that $Ker(\phi)≤20$. That rules out both the complete group $S_5$ and the alternating group $A_5$, so $Ker(\phi)$ must be trivial. As was pointed out (correctly) in the question this is all that was needed to complete the argument.

$\endgroup$
2
$\begingroup$

$S_6$ has two conjugacy classes of subgroups isomorphic to $S_5$. First of all the "trivial" one: stabilizers of a point. That gives you 6 subgroups. The second is the conjugacy class of $\langle(1234),(3456)\rangle$. Again $6$ subgroups. The subgroups in these two classes are not conjugate in $S_6$, because the members of the first fix a point, while the members of the second one act transitively.

$\endgroup$

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.