Are the conjugacy classes normal subgroups in $S_n$?

Question

I am working on proving that if $C$ is a conjugacy class in $S_n$ then $C^g =C$ for all $g \in S_n$.

I went over similar questions which only prove that a normal subgroup is contained in union of conjugacy classes, but not claiming a conjugacy class is normal subgroup.

A singe conjugacy class in $S_n$ is not a subgroup except for the trivial group, because it doesn't contain the neutral element $id$. — Dietrich Burde, Feb 07 '21 at 17:03
@DietrichBurde but proving that $C^g =C$ for all $g$ doesn't imply that $C$ is normal? — math_for_ever, Feb 07 '21 at 17:06
You wrote "normal subgroup" in the question. It is not a subgroup. But the word "normal" is only defined for subgroups in this context - or did you have a definition of a "normal set"? — Dietrich Burde, Feb 07 '21 at 17:06

score 1 · Answer 1 · answered Feb 08 '21 at 05:17

We don’t talk about “normality” except for subgroups. You may ask whether a set is “closed under conjugation”, and conjugacy classes certainly are: let $C$ be a conjugacy class, and let $x\in G$. To show $C^x\subseteq C$, let $g\in C$. Then $x^{-1}gx$ is conjugate to $g$, hence it lies in the conjugacy class of $g$... that is, in $C$. Thus, $x^{-1}Cx\subseteq C$. As this holds for all $x\in G$, it follows that $C^g=C$ for all $g\in G$.

Note that this holds for any group, not just $S_n$>

score 1 · Answer 2 · answered Mar 13 '21 at 12:55

Let $G$ be any group, $a\in G$ and $\operatorname{cl}(a):=\{g'ag'^{-1}, g'\in G\}$. Then:

\begin{alignat}{1} g\operatorname{cl}(a)g^{-1} &= \{gg'ag'^{-1}g^{-1}, g'\in G\} \\ &= \{(gg')a(gg')^{-1}, g'\in G\} \\ \tag 1 \end{alignat}

Since, for every $g\in G$, the map $g'\mapsto gg'$ is onto $G$, from $(1)$ follows: $g\operatorname{cl}(a)g^{-1}=\{g''ag''^{-1}, g''\in G\}=\operatorname{cl}(a)$, for every $g\in G$.

Are the conjugacy classes normal subgroups in $S_n$?

2 Answers2