Is $H$ a characteristic subgroup of $G$ when $G$ is abelian?

[Chris L T521]

Thanks to those who participated in last week's POTW! Here's this week's problem!

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Problem: Let $G$ be a finite group and for some fixed $n\in\mathbb{N}$ let $H=\{g\in G: g^n=1\}$. If $G$ is abelian, prove that $H$ is a characteristic subgroup of $G$ (Recall that $H$ is a characteristic subgroup of $G$ if (i) $H\leq G$ and (ii) for every $f\in\text{Aut}(G)$, $f(H)=H$). If $G$ isn't abelian, find a specific $G$ and $n\in\mathbb{N}$ that shows $H$ is not a characteristic subgroup in $G$ (in this case, it just suffices to show that $H$ is not a subgroup of $G$).

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[Chris L T521]

This week's question was correctly answered by Sudharaka and Deveno. You can find Sudharaka's solution below.

Spoiler

We shall first show that \(H\leq G\). Take any two elements \(h_1,h_2\in H\). Then, \(h_{1}^{n}=1\) and \(h_{2}^{n}=1\). Note that, \((h_{2}^{n})^{-1}=(h_{2}^{-1})^{n}=1\). Therefore,

\[h_{1}^{n}(h_{2}^{-1})^{n}=1\]

Since \(G\) is Abelian,

\[(h_{1}h_{2}^{-1})^{n}=1\]

\[\Rightarrow h_{1}h_{2}^{-1}\in H\mbox{ for each }h_1,h_2\in H\]

\[\therefore H\leq G\]

Let, \(f\in\text{Aut}(G)\) and take any \(h\in H\). Then,

\[[f( h)]^{n}=f(h^{n})\]

Since \(h^{n}=1\) we have,

\[[f( h)]^{n}=f(1)=1\]

\[\therefore f( h)\in H\mbox{ for each }h\in H\]

That is,

\[f(H)\subseteq H~~~~~~~~~~~~(1)\]

Since \(f\) is an automorphism it is surjective. Therefore for each \(h\in H\) there exist \(g\in G\) such that,

\[h=f(g)\]

\[\Rightarrow h^n=[f(g)]^n=f(g^n)\]

\[\Rightarrow f(1)=f(g^n)\]

Since f is injective,

\[g^{n}=1\]

\[\Rightarrow g\in H\]

\[\therefore h\in f(H)\mbox{ for each }h\in H\]

\[\Rightarrow H\subseteq f(H)~~~~~~~~~~~~(2)\]

By (1) and (2),

\[f(H)=H\]

Take the symmetric group of three symbols, \(S_{3}\) and let \(n=2\). \(S_{3}\) is a non-Abelian group and \(H=\{s\in S_{3}:s^{2}=1\}=\{(1),(1\ 2),(1\ 3),(2\ 3)\}\). But \(H\) is not a subgroup of \(S_{3}\).