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.

Hint: $G$ maps to the automorphisms of $H$ by conjugation. What is that automorphism group? — lulu, Jun 20 '23 at 11:40
Isn't $G=S_3, H=A_3$ a counter-example? If $a$ is an even permutation and $g$ is any permutation, then $gag^{-1}$ is even. But $(1\ 2\ 3) \in A_3$ and $(1\ 2\ 3)(1\ 2) = (1\ 3)$ but $(1\ 2)(1\ 2\ 3) = (2\ 3)$. — aschepler, Jun 20 '23 at 12:32
I understand both the answers ,which suggest an alternate way to proof this,but is there any way to complete my attempt, means what is the contradiction,if I assume $ga = a^2g$.please help — ビキマンダル, Jun 20 '23 at 16:23

Kan't · Answer 1 · 2023-06-20T13:45:07.750

5

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 $H\le Z(G)$.

edited Jun 20 '23 at 13:45

answered Jun 20 '23 at 13:10

Kan't

4,819

"it is the union of conjugacy classes" What does the conjugacy class look like? Can you write it down explicitly and symbolically? – Jun 20 '23 at 15:18
1

See e.g. here, @GGplay: https://math.stackexchange.com/q/2175828/1092170 – Kan't Jun 20 '23 at 15:30
What's the downvote for? – Kan't Jun 20 '23 at 15:35
Suppose $a\neq e$, if $[a]={ gag^{-1}|\forall g\in G}$, why $|[a]|\neq 2$? Because $[a]$ is NOT a subgroup, there is no contradiction with $|G|$ is odd. – Jun 20 '23 at 16:47
Conjugacy classes are orbits, and orbits' sizes divide the order of the group (orbit-stabilizer), @GGplay. – Kan't Jun 20 '23 at 16:49
i see, thank you! Can you add this to the answer? otherwise the system won't let me remove the downvote :) – Jun 20 '23 at 16:52
1

You shouldn't downvote just because you don't get some parts. – Kan't Jun 20 '23 at 17:28
2

No, when you wrote an answer you need to make it clear to other readers, not only for expert, because expert doesn't need your answer. In your answer you use two important results without mentioning it, which is difficult to follow. – Jun 20 '23 at 18:22
1

It's impossible to state a priori what "other readers"(?) know/don't know, @GGplay. With this unit of measure, every answer ought to be endlessly self-consistent. I made a choice, and I'll get stick to it. No problem, though. – Kan't Jun 20 '23 at 18:27
1

An excellent answer, @citadel, using only basic facts that anyone asking this question should know. The notion that anything you've used is "expert" stuff is nonsense. – ancient mathematician Jun 21 '23 at 08:04

SoG · Answer 2 · 2023-06-21T07:16:15.167

2

Alternative:

Hints:

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

$$\begin{align}N_G(H) /C_G(H) &\cong K\\&\le \textrm{Aut(H)}\end{align}$$
$H\lhd G$ implies $N_G(H) =G$
$\textrm{Aut}(H) \cong \Bbb{Z}_2$
$|G /C_G(H)|\neq 2$ as $|G|=\textrm{odd}$

edited Jun 21 '23 at 07:16

answered Jun 20 '23 at 15:19

SoG

13,327

How to prove the 1st result? Please explain – ビキマンダル Jun 21 '23 at 07:12
Define $\phi: N_G(H) \to \textrm{Aut}(H) $ by $$\phi(g) =\phi_g$$ where $\phi_g(x) =g^{-1}xg$ – SoG Jun 21 '23 at 09:29
$\phi$ is homomorphism. $$\$$ 2) $\ker(\phi) =C_G(H) $

SoG

Jun 21 '23 at 09:31

score 2 · Accepted Answer · answered Jun 21 '23 at 17:09

2

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).

answered Jun 21 '23 at 17:09

Kan't

4,819

1

Oh right.Thank you – ビキマンダル Jun 21 '23 at 17:18

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

3 Answers3