Conjugation in $A_n$

Definition: Let $G$ be a group and $a,b\in G$. We say that $b$ is a conjugate of $a$ if $b=cac^{-1}$ for $c\in G$.

It's easy to verify that conjugation is an equivalence class in $G$.

Also, we know the following fact: $\sigma, \tau\in S_n$ are conjugate in $S_n$ $\Leftrightarrow$ $\sigma, \tau$ have the same cycle decomposition.

Let's consider the normal subgroup $A_n$ instead of $S_n$. We know that $A_n$ is generated by $3$-cycles.

Theorem: If $n\geq 5$ we can prove that $3$-cycles are conjugate in $A_n$.

Let's consider $n<5$. Firstly, let's deal with $A_3$ and take $(123)$ and $(132)$. Even though $(123)$ and $(132)$ are $3$-cycles, they are not conjugare since $(123)\neq\theta(132)\theta^{-1}$ for any $\theta \in A_3=\{e, (123), (321)\}$. Thus, the theorem fails for $n=3$.

Let's take the group $A_4$. In one lecture (namely, Keith Conrad's lecture) I have read the following: Let's take $(123)$ and $(132)$. Although they are $3$-cycles, $(123)=(32)(132)(32)$ hence they are not conjugate.

How does it follow that they are not conjugate? What if for some $\theta\in A_4$ we get $(123)=\theta (132)\theta^{-1}$?

Would be grateful if somebody could explain my question, please.

• See for example here. The list of Related questions in the right margin has other promising hits as well. – Jyrki Lahtonen Jan 20 '18 at 17:10
• In the case of $A_4$ you can also argue as follows. There are a total of $8$ 3-cycles in $A_4$. But $8$ is not a factor of $|A_4|=12$. Because the size of a conjugacy class must always be a factor of the order of the group it follows that not all 3-cycles can be conjugates in $A_4$ (even though they are conjugate in $S_4$). In fact the 3-cycles split into two conjugacy classes of four each (see the linked question or another related one). – Jyrki Lahtonen Jan 20 '18 at 17:14
• @JyrkiLahtonen, I would like to understand why if $(123)=(32)(132)(32)$ how it follows that $(132)$ and $(123)$ are not conjugate? Could you explain it, please? – ZFR Jan 20 '18 at 18:04
• Oh, sorry. In $S_4$ the 3-cycle $(123)$ as eight conjugates. Therefore it is centralized by $24/8=3$ elements of $S_4$. It then follows that the centralizer $C$ of $(123)$ in $S_4$ consists of the powers $(123)^k, k=0,1,2$. We see that $(23)(123)(23)=(132)$. This implies that all the permutations $\sigma\in S_4$ such that $\sigma(123)\sigma^{-1}=(132)$ are the elements of the form $(23)\tau,\tau\in C$. Here the elements of $C$ are all even permutations. Hence the permutations $(23)\tau$ are oll odd. – Jyrki Lahtonen Jan 20 '18 at 18:35
• @JyrkiLahtonen, What does mean the phrase "it is centralized by"? – ZFR Jan 20 '18 at 19:08

Consider the action of $A_4$ on itself and the associated conjugacy classes.

Let $(u,v,w) \in A_4$. Then, $(u,v,w)^{-1} = (w,v,u)$.

Suppose $\exists \tau \in A_4: (\tau(u),\tau(v),\tau(w)) = \tau\circ (u,v,w)\circ \tau^{-1} = (w,v,u) = (u,w,v)$.

Then, $\tau(u) = u$ and $\tau(v) = w$ and $\tau(w) = v$, that is, $\tau \in \{(1,2)\circ (3,4), (1,3)\circ (2,4),(1,4)\circ (2,3)\}$ which is a contradiction to $\tau(u) = u$.

Therefore, $(u,v,w)$ and $(u,v,w)^{-1}$ are in different conjugacy classes in $A_4$.

EDIT: $A_4$ has the following conjugacy classes

• $\{\iota\}$
• $\{(1,2)\circ (3,4), (1,3)\circ (2,4),(1,4)\circ (2,3)\}$
• $\{(1,2,3),(2,4,3),(1,3,4),(1,4,2)\}$
• $\{(1,3,2),(2,3,4),(1,4,3),(1,2,4)\}$

What I was trying to argue above is that the inverse $(w,v,u)$ of any $3$-cycle $(u,v,w)$ in $A_4$ lies in another conjugacy class than $(u,v,w)$.

So, I suppose (wrongly) that $(w,v,u)$ and $(u,v,w)$ lie in the same conjugacy class, that is, there is a permutation $\tau$ in $A_4$ which maps $(w,v,u)$ to $(u,v,w)$. Iny symbols: $\tau\circ (u,v,w)\circ \tau^{-1} = (w,v,u)$.

We know $(\tau(u),\tau(v),\tau(w)) = \tau\circ (u,v,w)\circ \tau^{-1}$. If you look really closely at $(\tau(u),\tau(v),\tau(w)) = \tau\circ (u,v,w)\circ \tau^{-1} = (w,v,u)$ you will see that $\tau$ has a fixed point, that is, either $\tau(u) = u$ or $\tau(v) = v$ or $\tau(w) = w$. I chose $\tau(u) = u$.

Moreover, $\tau$ contains a transposition: $\tau(v) = w$ and $\tau(w) = v$ by my choice $\tau(u) = u$. The only permutations in $A_4$ which contain a transposition are $\{(1,2)\circ (3,4), (1,3)\circ (2,4),(1,4)\circ (2,3)\}$. But none of these permutations has a fixed point. Contradiction!

Does my argument make more sense to you now?

• Honestly, I did not understand your proof. What arr you proving? Could you expand your answer a bit? – ZFR Jan 20 '18 at 17:58
• @RFZ: I will edit my answer. – Moritz Jan 20 '18 at 20:56