Prove that $a$ commutes with every element of the group $G$.

Question

Let $G$ be a finite group having an odd number of elements. Suppose $a$ is an element of $G$ of order $3$ such that the cyclic subgroup $ H$ generated by $a$ is normal in $G$. Prove that $a$ commutes with every element of $G$.

My attempt: $|a| = 3$,

so $H= \{ e, a,a^2\}$

Let $g \in G$ be arbitrary element.

Now $H$ is normal in $G$, so $gag^{-1} \in H$

In particular, $ga\in \{g,ag,a^2g\}.$

If $ga = g$, then $a=e$ (contradiction)

If $ga = ag$, then $a$ commutes with $g$, so we are done.

If I can show that $ ga \ne a^2g$ then we get our required result. I think I need to use the order of the group, but couldn't understand how to use that.

So any hint/solution to show $ ga \ne a^2g$ or any better way to solve this problem will be helpful for me.

Thanks in advance.

Answer 1

Your argument works, and can be completed as follows: $ga=a^2g$ if and only if $gag^{-1}=a^2$, meaning that $a$ and $a^2$ are in the same conjugacy class, which must then be fully contained in the normal subgroup $\langle a\rangle=\{e, a, a^2\}$. But then this conjugacy class has size $2$, which is impossibile because $G$ has odd order and classes' sizes divide group's order (orbit-stabilizer).

Answer 2

Since $H=\langle a\rangle$ of order $3$ is normal, it is the union of conjugacy classes. Since $|G|$ is odd, there isn't any conjugacy class of size $2$, and hence necessarily $H$ is the union of three singleton classes, namely $H\le Z(G)$.

Answer 3

Alternative:

Hints:

Let $G$ be a group and $H\le G$.

1. $$\begin{align}N_G(H) /C_G(H) &\cong K\\&\le \textrm{Aut(H)}\end{align}$$

2. $H\lhd G$ implies $N_G(H) =G$

3. $\textrm{Aut}(H) \cong \Bbb{Z}_2$

4. $|G /C_G(H)|\neq 2$ as $|G|=\textrm{odd}$