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If $H \triangleleft G$ and $|G/H|=m$, show that $x^m \in H$ for $\forall x \in G$.

My attempt is: since $|G/H|=\frac{|G|}{|H|}=m$, we have $x^{|G|}=x^{m|H|}=e$, then I stuck here. Can anyone guide me ?

Idonknow
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  • The quotient group $G/H$ has order $m$. In a group of order $m$, we have $x^m=e$ for all $x$. Now in this case, replace $x$ by $xH$ and $e$ by $H$. – Julien Mar 07 '13 at 04:26
  • More generally, your result holds as long as $|G:H|=m$. For otherwise, there is $x\in G\setminus H$ such that $o(x)>m$. By counting the number of cosets of $H$, you will find that $\langle H,x\rangle$ will give a group that have larger size than $G$. – Easy Mar 07 '13 at 06:22

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Since $(xH)^m=H$, it follows that $x^mH=H$, and thus that $x^m\in H$.

Ittay Weiss
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Your given that $H \triangleleft |G/H| = m.\,$ Hence, for every $\,x\,$ in a group of order of order $m$, $x^m = e$.
For the group $\,G/H\,$, this means that $\,(xH)^m = H\,$ for all $x \in G.\;\;$ Why?

But as $\,H\,$ is normal in $\,G,\;$ $(xH)^m = H \quad\implies\quad x^mH = eH = H\quad \implies x^m \in H$.

amWhy
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