$S_6$ contains two subgroups that are isomorphic to $S_5$ but are not conjugate to each other

Question

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?

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. — lulu, Aug 04 '15 at 02:32
@lulu Why do you say $Ker\phi$ has order no greater than 20, then it is trivial? — AG learner, Aug 04 '15 at 02:41
Because $Ker\varphi$ is normal. $S_5$ hasn't got any non-trivial normal subgroups further down than index $2$. — lulu, Aug 04 '15 at 02:43
Side note: I was asked this question at my Phd orals, some while back. — lulu, Aug 04 '15 at 02:45
@lulu Haha, it's a coincidence. On which year should a Ph.D. student has his/her oral exam, or it is a part of exam of Algebra course? — AG learner, Aug 04 '15 at 02:49

score 11 · Accepted Answer · answered Aug 05 '15 at 10:17

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.

score 7 · Answer 2 · answered Aug 05 '15 at 08:42

$S_6$ 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.

suckling pig · Answer 3 · 2024-07-26T21:25:28.733

2

Transitivity of a subgroup will be preserved by conjugation, since the stabilisers will have the same size (the conjugate of the stabiliser is the stabiliser of the conjugate; check that).

There's $6$ obvious non-transitive $S_5$'s in $S_6.$

There's also famously a transitive $S_5,$ corresponding to an outer automorphism.

Therefore they're not conjugate.

edited Jul 26 '24 at 21:25

answered Jul 14 '24 at 09:23

suckling pig

1

score 2 · Answer 4 · answered Jul 27 '24 at 03:29

2

Another approach is to use $S_5\cong PGL(2,5)$, which acts faithfully and (sharply 3-)transitively on the projective line over $\mathbb{F}_5$, having 6 points. This gives an embedding as a transitive subgroup of $S_6$.

answered Jul 27 '24 at 03:29

Steve D

5,737

$S_6$ contains two subgroups that are isomorphic to $S_5$ but are not conjugate to each other

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