Show that $\langle S\rangle$ is a normal subgroup of $G$.

Question

Let $S$ is a subset of a group $G$, and $gsg^{-1} \in S$ for each $s \in S$ and $g \in G$. Show that $\langle S\rangle$ is a normal subgroup of $G$.

I am stuck on this problem. Is anyone is able to give me a hint?

Answer 1

Hint: Let $\langle S \rangle$ be the subgroup of $G$ generated by all products of $s \in S$ and their inverses. Note that for $s_1, s_2$ and a particular $g$ in $G$, we have $s_1s_2 = s_1g^{-1}gs_2$.

Answer 2

Here is slightly different approach which uses the characterization that $\langle S \rangle$ is the smallest subgroup of $G$ which contains $S$. In other words, if $H$ is a subgroup of $G$ with $S \subseteq H$, then $\langle S \rangle \subseteq H$.

Let $gSg^{-1}$ denote the set $\{gsg^{-1} \mid s \in S\}$. We are given that $gSg^{-1} \subseteq S$, or equivalently $S \subseteq g^{-1}Sg$, for every $g \in G$.

Now, since $S \subseteq \langle S \rangle$, we have $S \subseteq g^{-1}Sg \subseteq g^{-1}\langle S \rangle g$. This means that $g^{-1}\langle S \rangle g$ is a subgroup of $G$ which contains $S$. But $\langle S \rangle$ is the smallest such subgroup, so $\langle S \rangle \subseteq g^{-1}\langle S \rangle g$. Equivalently, $g\langle S \rangle g^{-1} \subseteq \langle S \rangle$. This holds for every $g \in G$, so $\langle S \rangle \lhd G$.