Are there two isomorphic transitive subgroups of $\mathfrak{S}_n$ that are not conjugate?

Question

I know that we can find two subgroups of $\mathfrak{S}_6$ both isomorphic to $\mathfrak{S}_5$ that are however not conjugate (here) in $\mathfrak{S}_6$. These subgroups are not conjugate precisely because one and only one of them is transitive.

Can we find two transitive subgroups of a symetric group that are isomorphic but not conjugate ?

See this post, item 2. – Dietrich Burde Jun 10 '22 at 07:58 — Dietrich Burde, Jun 10 '22 at 07:58

Derek Holt · Answer 1 · 2022-06-10T08:13:08.660

4

You are looking for a group with two core-free subgroups of the same order that are not equivalent under an automorphism of the group.

For example, we could take the subgroups $\langle (1,2) \rangle$ and $\langle (1,2)(3,4) \rangle$ of $S_4$. The actions on the cosets of these subgroups give rise to two transitive subgroups of $S_{12}$ that are isomorphic to $S_4$, but not conjugate in $S_{12}$.

Or, we could take the subgroups $\langle (1,2,3,4) \rangle$, $\langle (1,2),(3,4) \rangle$ of $S_4$ to get an example in $S_6$.

edited Jun 10 '22 at 08:13

answered Jun 10 '22 at 08:01

Derek Holt

90,008

Ok but why these two transitive subgroups of $S_{12}$ are then not conjugate ? – Antoine de La Humâlerie Jun 10 '22 at 08:54
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If they were conjugate, conjugation would map one point stabilizer isomorphically to the other one. – ahulpke Jun 10 '22 at 10:53

Are there two isomorphic transitive subgroups of $\mathfrak{S}_n$ that are not conjugate?

1 Answers1