Let $G$ be a group and $g \in G$ have order $2$. Show that $\langle g \rangle \unlhd G \iff g \in Z(G)$.

Question

This is my proof attempt, if something is wrong, then please correct it.

$\implies$ Let $g^* \in G$. If $\langle g \rangle \unlhd G$, then $g^* \langle g \rangle = \langle g \rangle g^*$. So $\{g^*g, g^*\} = \{gg^*, g^*\}$. From that $g^*g = gg^*$, so $g \in Z(G)$

$\impliedby$ Let $g_* \in G$. If $g \in Z(G)$ then $g_*g = gg_*$. Moreover $g_*g^2 = g_* = g^2g_*$. So $g_*\langle g \rangle = \langle g \rangle g_*$. Therefore $\langle g \rangle \unlhd G$.

Thanks.

Answer 1

Yes, your proof is correct. Well done!

One thing worth pointing out, though, is that the $\Rightarrow$ and the $\Leftarrow$ symbols are best put between parentheses. But this is just a matter of style.