# Indicate two nonconjugate isomorphic subgroups of order $12$ in $S_{11}$

To my understanding, two subgroups are conjugates if they belong to the same conjugacy class. So I have to ''pick'' subgroups from different conjugacy classes. Of course the chosen subgroups have to be of the same order in order to be isomorphic.

But how may I go about coming up with those groups? Any hint would be helpful.

• Think about S3 with (123) and with (123)(456) – Steve D Apr 2 '17 at 18:38
• Or you could try writing down two elements of order $12$ that have different cycle types. Then the groups they generate will not be conjugate. – Derek Holt Apr 2 '17 at 18:47
• @DerekHolt and how do I go about proving that they are isomophic? – Jazz Apr 2 '17 at 18:53
• Do the same as in this question. – Dietrich Burde Apr 2 '17 at 18:56
• Note that there is a difference between conjugacy classes of elements and conjugacy classes of subgroups. In general, a subgroup will not be contained inside a conjugacy class of elements. – arctic tern Apr 2 '17 at 19:10

If $H\le G$ is a subgroup and $g\in H$ an element, then $gHg^{-1}=\{ghg^{-1}:h\in H\}$ (the image of $H$ under the automorphism $f(x):=gxg^{-1}$) is a conjugate of $H$.

(Usually the term "conjugacy class" refers to "conjugacy class of elements," in which case the conjugacy class of $x$ is given by $\{gxg^{-1}:g\in G\}$. A conjugacy class will never contain a subgroup, except for the conjugacy class $\{e\}$ which is itself the trivial subgroup.)

For permutations, conjugating a permutation preserves its cycle type. (In fact, the conjugacy class of a permutation - the set of all other permutations conjugate to it - will be the set of all permutations with the same cycle type.) Thus, conjugating a subgroup will not change the number of permutations it has of each cycle type.

So, here's an idea. Find two different elements of order $12$ that have different cycle types. See if the subgroups they generate have the same number of elements of each cycle type.

• I was considering $\tau = (123)(4567)$ and $\sigma = (12)(345)(7689)$. We have that $o(\tau) = \mathbb{lcm}(3,4) = 12 = \mathbb{lcm}(2, 3,4) = o(\sigma)$. Since is we take the cyclic subgroup $(\langle \tau \rangle)$ and $(\langle \sigma\rangle)$ we have that they are isomorphic since they have the same order. Am i mistaking somewhere here? – Jazz Apr 2 '17 at 19:53
• @Jazz Nope, no mistake. Now does $\langle\tau\rangle$ have any elements with the same cycle type as $\sigma$? – arctic tern Apr 2 '17 at 19:54
• So if I'm understanding: I have to ''take power'' of $\tau$ and $\sigma$ up to $11$ and check whether $\tau^i \neq \sigma^j$ for $i,j \in (1,11)$ (in the sense of cycle type). Only if this condition holds for every $i$ and $j$ in $(1,11)$ I can claim that $(\langle \tau \rangle)$ and $(\langle \sigma\rangle)$ are ''fully'' nonconjugates, right? – Jazz Apr 2 '17 at 20:04
• No, it's not enough to check the $\tau^i$s don't equal the $\sigma^j$s. That would only mean the two subgroups intersect trivially. To show they are not conjugate, it suffices to show none of the powers of $\tau$ have the same cycle type as $\sigma$. (Because if $\langle\tau\rangle$ and $\langle\sigma\rangle$ were conjugate, then $\langle\tau\rangle$ would contain an element conjugate to $\sigma$, i.e. of the same cycle type. You can prove that right?) – arctic tern Apr 2 '17 at 20:07
• No, taking powers of $\sigma$ is not necessary. – arctic tern Apr 2 '17 at 20:19