If $\mathrm{Stab}(x)_{S_5} \nsubseteq A_5$ then how to prove $[\mathrm{Stab}(x)_{S_5} : \mathrm{Stab}(x)_{A_5}]=2$? Please help

Question

I am trying to calculate the conjugacy classes of $A_5$.

The conjugacy classes of $A_5$ are the orbits of the action of $A_5$ in $A_5$ given by the conjugacy action. You know that in $S_5$ every one of those elements is in a unique conjugacy class, and they represent all the classes. You know from the class equation that (in $S_5$)$$|\mathrm{orb}(x)_{S_5}|=\frac{|S_5|}{|\mathrm{Stab}(x)_{S_5}|}$$

We are trying to study $\mathrm{orb}(x)_{A_5}$.

Since $\mathrm{Stab}(x)_{S_5} < S_5$, we have two possibilities:

1) $\mathrm{Stab}(x)_{S_5} \subseteq A_5$: in this case $|\mathrm{orb}(x)_{A_5}|=\frac{1}{2}|\mathrm{orb}(x)_{S_5}|$, so your class in $S_5$ splits in two new classes in $A_5$.

2) $\mathrm{Stab}(x)_{S_5} \nsubseteq A_5$: since $A_5$ is a subgroup of index 2, and since $$\mathrm{Stab}(x)_{A_5}=A_5 \cap \mathrm{Stab}(x)_{S_5}$$ If I get$$[\mathrm{Stab}(x)_{S_5}:\mathrm{Stab}(x)_{A_5}]=2$$ we have $|\mathrm{orb}(x)_{A_5}|=|\mathrm{orb}(x)_{S_5}|$, and we get the same conjugacy class.

Moral: you just have to know if $\exists \tau \in \mathrm{Stab}(x)_{S_5}$ such that $\tau \notin A_5$

Now I am not able to prove the 2nd part $$[\mathrm{Stab}(x)_{S_5}:\mathrm{Stab}(x)_{A_5}]=2$$.

Please help me in here.

Was I clear?

Answer 1

Let us get the information of the conjugacy classes in $S_5$ and $A_5$, here using sage, so that we have a clear situation.

sage: S5 = SymmetricGroup(5)
sage: S5.order()
120
sage: S5.conjugacy_classes()
[Conjugacy class of cycle type [1, 1, 1, 1, 1] in Symmetric group of order 5! as a permutation group,
 Conjugacy class of cycle type [2, 1, 1, 1] in Symmetric group of order 5! as a permutation group,
 Conjugacy class of cycle type [2, 2, 1] in Symmetric group of order 5! as a permutation group,
 Conjugacy class of cycle type [3, 1, 1] in Symmetric group of order 5! as a permutation group,
 Conjugacy class of cycle type [3, 2] in Symmetric group of order 5! as a permutation group,
 Conjugacy class of cycle type [4, 1] in Symmetric group of order 5! as a permutation group,
 Conjugacy class of cycle type [5] in Symmetric group of order 5! as a permutation group]
sage: A5 = AlternatingGroup(5)
sage: A5.order()
60
sage: A5.conjugacy_classes()
[Conjugacy class of () in Alternating group of order 5!/2 as a permutation group,
 Conjugacy class of (1,2)(3,4) in Alternating group of order 5!/2 as a permutation group,
 Conjugacy class of (1,2,3) in Alternating group of order 5!/2 as a permutation group,
 Conjugacy class of (1,2,3,4,5) in Alternating group of order 5!/2 as a permutation group,
 Conjugacy class of (1,2,3,5,4) in Alternating group of order 5!/2 as a permutation group]

Now i try to answer the question, the strategy of finding the conjugacy classes is not very clear, let's see if i am using the same idea.

Fix some $x\in S_5$. Its (cycle decomposition) type already characterizes the conjugacy class in $S_5$. So $sxs^{-1}$ covers for a running $s\in S_5$ all permutations of same type.

Now let us start with an $x\in A_5$. Among the allowed types are $11111$, $221$, $311$, $5$. We try to see when the $S_5$-related type / $S_5$-conjugacy class splits when considering the smaller action by conjugation of $A_5$.

• The stabilizer in $S_5$ for the conjugacy action of the element $(12345)\in A_5$ (i.e. its commutator) is the cyclic group generated by it, so it is included in $A_5$, using (1) in the OP we see that the conjugacy class splits in two classes,
  represented by $(12345)$ and $(12354)$.

• Now let us consider $x\in A_5$ with type $311$. For example $(123)(4)(5)=(123)$, then $(45)$ fixes it, commutes with it. In general, for $(abc)(d)(e)$ we use the odd permutation $(de)$ to see that the stabilizer in $S_5$ is "double" of the one in $A_5$.

• Now let us consider $x\in A_5$ with type $221$. For example $(12)(34)(5)=(12)(34)$, then the odd $(12)$ fixes it, commutes with it. In general, for $(ab)(cd)(e)$ we use the odd permutation $(ab)$ to see that the stabilizer in $S_5$ is "double" of the one in $A_5$.

• The case of the conjugacy class of the neutral element is clear.

Answer 2

Use Lagrange’s theorem and the 2nd isomorphism theorem for groups $H/N \cap H \cong NH/N$ with $N=A_5 \unlhd S_5$ and $H=Stab(x)_{S_5} \leq S_5$ to prove $A_5 \cdot Stab(x)_{S_5}=S_5$. With $\mathrm{Stab}(x)_{A_5}=A_5 \cap \mathrm{Stab}(x)_{S_5}$ the claim then follows.