Conjugation in $S_4$: orbit and stabilizers.

Question

Good morning all.

By operating $S_4$ on itself by conjugation, what will be its orbits and stabilizers?

Here is what I tried to do: I listed the $24$ elements of $S_4$ but I think that combining the elements between them would be a bit long. I need directions

Answer 1

There is a very nice general formula for conjugation in $S_n$, which is $$ \sigma \: (a_1 \cdots a_k)(b_1 \cdots b_l) \cdots \sigma^{-1} = (\sigma(a_1) \cdots \sigma(a_k))(\sigma(b_1) \cdots \sigma(b_l)) \cdots, $$ assuming the cycles between $\sigma$ and $\sigma^{-1}$ are all disjoint. This means that we can compute conjugates easily by considering the decomposition in disjoint cycles.

By the way, the proof of this formula is not very hard: you just show $\sigma(a_i)$ is mapped to $\sigma(a_{i+1})$, and similarly for $b_i$, and all other cycles in the cycle decomposition.

Answer 2

There is a theorem that cycle structure is preserved iff the permutations are conjugate. See the first answer.

In $S_4$ the possible cycle structures are $(),(abcd),(abc), (ab)(cd)$ and $(ab)$, since there is also a theorem that any permutation can be written as a product of disjoint cycles.

Now there are ${4\choose2}=6$ transpositions ($2$-cycles). $3$ products of two transpositions. $4!/4=6 $ $4$-cycles. And $8$ $3$-cycles. And finally the identity forms its own orbit.

The class equation sums this up: 1+3+6+6+8.

Now you can get the orders of the stabilizers from the orbit-stabilizer theorem. To get the actual stabilizers you may need to do it by hand.