# Problem of the Week #33 - November 12th, 2012

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#### Chris L T521

##### Well-known member
Staff member
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

##### Well-known member
Staff member
This week's question was correctly answered by Sudharaka and Deveno. You can find Sudharaka's solution below.

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 $$S_{3}$$.

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