As part of a larger proof, I am proving that $S=\{s\in G \mid s=s^{-1} \}$ is a normal subgroup with respect to some group $G$. If I can prove this (seemingly simple) fact, the rest of my proof falls into place quite well.
I tried to show that $gsg^{-1}\in S$ for $g\in G$ but was unsuccessful. Without solving this problem, could anyone provide me with a helpful hint, nudging me in the right direction?
Let $G$ be a group and $n$ be an arbitrary positive integer, and let $S=\langle s \in G: s^n=1 \rangle$, the group generated by all elements of $G$ of which the order divides $n$. Then $S$ is a normal subgroup. By definition $S$ is a subgroup, so let $g \in G$ and $s \in S$. Then $(g^{-1}sg)^n=g^{-1}sg \cdot g^{-1}sg \cdots g^{-1}sg$ ($n$ times)$=g^{-1}s \cdot s \cdots sg=g^{-1}s^ng=g^{-1}g=1$. Hence $g^{-1}sg \in S$ and $S$ is normal. For $n=2$ this yields your post, Ethan.