Let $G$ be a group of order $16$ with an element $g$ of order $4$. Prove that the subgroup of $G$ generated by $g^2$ is normal in $G$. Question: Let $G$ be a group of order $16$ with an element $g$ of order $4$.  Prove that the subgroup of $G$ generated by $g^2$ is normal in $G$.
Thoughts: I keep getting stuck.  My most recent path is that since $G$ is solvable, then there is a subnormal series... so maybe I can get something to come out there?  But I just feel like a counterexample is throwing a wrench into everything I try.  Any help would be greatly appreciated.
 A: Here is a hint. Let $H=\langle g\rangle$. Now, consider the action of $G$ on the cosets of $H$. Since $|G:H|=4$, this gives you a homomorphism from $G$ to $S_4$. Since $16$ does not divides $|S_4|$, this homomorphism cannot be injective. What can you say about the kernel?
A: Verret's solution is very nice, and surely how it is meant to be done. But here is another way:
Note that we aim to prove that $g^2$ is central, so assume not. Let $H=C_G(g)$. If $H=G$ then $g\in Z(G)$, so $g^2\in Z(G)$, a contradiction. Since $Z(G)$ is non-trivial and contained in $H$, and does not contain $g^2$, $H$ must be $\langle g\rangle \times Z(G)$ (with $|Z(G)|=2$). (This is purely because $Z(G)$ centralizes $g$ and so forms a direct product.) But then $H$ is normal, so $G$ normalizes the set of squares of elements of $H$, i.e., $\{1,g^2\}$. Hence $H^2$ is a normal subgroup of $G$, and done.
Edit: Here's another proof, using a fact that everyone should know: any two elements of order $2$ generate a dihedral group.
Let $x=g^2$. Since $x$ is non-central, there is a conjugate $y$ of $x$ with $x\neq y$. Consider $H=\langle x,y\rangle$, a dihedral group. Since the reflections in dihedral groups aren't squares, $g\notin H$. Also, two generating reflections in a dihedral $2$-group, like $H$, are not conjugate in $H$. Thus $H\leq C_G(x)$. Thus, since $H$ and $g$ are contained in the proper subgroup $C_G(x)$, we have $|H|=4$, i.e., $x$ commutes with its conjugate $y$. Notice also that $g$ must commute with $y$ as well, so $\langle g,y\rangle=\langle g\rangle \times \langle y\rangle$. In this group, $y$ is not a square. But $y$ is conjugate to $x$, which is a square, so $y=h^2$ for some $h\in G$. But $h$ centralizes $x$ for the same reason that $g$ centralized $y$. Hence $C_G(x)> \langle g,y\rangle$, and is thus the whole group $G$.
