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Let $i_G(N)$ denote the index of the subgroup $N$ in $G$

This is an exercise from Herstein's "Topics in Algebra".

Let $N$ be a normal subgroup of a finite group $G$ such that $i_G(N)$ and $o(N)$ be relatively prime, show that any $x\in G$ satisfying $x^{o(N)}=e$ must imply that $x\in N$.

My first instinct is write out for some $a,b\in \mathbb Z$: $$ai_G(N)+bo(N)=1$$ so $$x=x^{ai_G(N)}$$ Since $a$ is unknown it seems like I should show that $$x^{i_G(N)}\in N$$ Why is this true?

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    Since $Nx\in G/N$, then $N=(Nx)^{i_G(N)}=N(x^{i_G(N)})$, and hence $x^{i_G(N)}\in N$. – Deif Apr 18 '25 at 20:01
  • Please delete this question since it is a dupe of a FAQ and there is nothing new here (so it will only clutter search results). You will first need to unaccept the answer. Please search for answers before asking questions. – Bill Dubuque Apr 19 '25 at 19:11

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We know that in a group if $x\in G , x^d = e, x\ne e$ then $d||G|$

So assume in the $G/N$ group that $Nx$ is not $N$, but from hypothesis we know that $(Nx)^{o(N) } = N$ so we got that : $o(N) |o(G/N) = i_G(N) $

But we know that: $\gcd(o(N),i_G(N))=1$ contradiction!

So we got that $x\in N$

Benyamin Khanzadeh H.
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  • Please strive not to post more (dupe) answers to dupes of FAQs (& PSQs), cf. site policy announcement here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe/PSQ processing.) – Bill Dubuque Apr 19 '25 at 19:10