## The subgroup generated by all elements of some fixed order is normal

Let $G$ be a group and let $a,b \in G$.

1. Let $g \in G$. Note that $g(ab)b^{-1} = gag^{-1} \cdot gbg^{-1}$. Prove that $|a| = |gag^{-1}|$.
2. Let $g \in G$. Prove that $ga^{-1}g^{-1} = (gag^{-1})^{-1}$.
3. Let $N = \langle S \rangle$ be a subgroup of $G$. Prove that if $gSg^{-1} \subseteq N$ for all $g \in G$, then $N$ is normal in $G$.
4. Prove that a cyclic subgroup $N = \langle x \rangle$ is normal in $G$ if and only if for each $g \in G$, $gx^{-1}g^{-1} = x^k$ for some $k \in \mathbb{Z}$.
5. Let $n$ be a positive integer. Prove that the subgroup $N \leq G$ generated by all elements of order $n$ is normal in $G$.

1. We have $g(ab)g^{-1} = ga1bg^{-1} = gag^{-1}gbg^{-1}$. Now by a lemma to a previous exercise, $(gag^{-1})^n = ga^ng^{-1}$ for all $n \in \mathbb{Z}$. Thus $|gag^{-1}| \leq |a|$. If $|gag^{-1}| = k$, then $ga^kg^{-1} = 1$, so that $a^k = 1$. Thus $|a| \leq |gag^{-1}|$.
2. Note that $gag^{-1}ga^{-1}g^{-1} = gaa^{-1}g^{-1} = gg^{-1} = 1$. By the uniqueness of inverses, $(gag^{-1})^{-1} = ga^{-1}g^{-1}$.
3. Let $N = \langle S \rangle$, and suppose $gSg^{-1}$ for all $g \in G$. Then for all $g \in G$, we have $gNg^{-1} = g \langle S \rangle g^{-1} = \langle gSg^{-1} \rangle$ $\subseteq \langle N \rangle = N$. Thus $N$ is normal in $G$.
4. Let $N = \langle x \rangle$. $(\Leftarrow)$ If $N$ is normal, then for all $g \in G$ we have $gxg^{-1} \in \langle x \rangle$. Thus $gxg^{-1} = x^k$ for some integer $k$. $(\Rightarrow)$ If for all $g \in G$ there exists an integer $k$ such that $gxg^{-1} = x^k \in N$, by the previous point we have $N$ normal in $G$.
5. Let $S \subseteq G$ consist of all elements of order $n$, and let $s \in S$. For all $g \in G$, by the first point we have $|gsg^{-1}| = |s| = n \in \langle S \rangle$. By the third point, $\langle S \rangle$ is normal in $G$.