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The question from the book:

Let $G$ be a finite group it's order $|G|=nm$, with $\gcd(m,n)=1$. Supposing the existence of a subgroup $H$ in $G$ such that $|H|=n$. Show that $H$ is the only subgroup in $G$ of order equal to $n$ if and only if $H$ is normal in $G$. (Hint: If $K$ is another subgroup of order $n$, consider $HK$.)

I have managed the first part, proving that if $H$ is unique than it is normal, but I'm having a lot of trouble proving the other.

I can see $HK$ is a subgroup of $G$, and that it's order will be $n^2/|H\cap K|$, but I'm not quite sure where to go from there. We have not yet learned Sylow's theorems, so I assume it must be possible without using them.

Martin Sleziak
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Rui Aldé Lopes
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    You have almost done it. If $H \ne K$ then $|H \cap K|$ is a proper divisor of $n$, so $|HK|$ does not divide $|G|$. – Derek Holt Oct 30 '20 at 07:23
  • @DerekHolt I can see it would be a proper divisor, but why won't |HK| divide |G|? Say |H∩K|.p=n and |G|=|HK|.q, can't |G|=|HK|.p.q=n.p.q? – Rui Aldé Lopes Oct 30 '20 at 07:36
  • Nevermind, that would mean m=p.q, but p|n, and so gcd(m,n)=/=1. Thank you. – Rui Aldé Lopes Oct 30 '20 at 07:43
  • Which book are you referring to? – Shaun Oct 30 '20 at 15:43
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    @Shaun I'm brazilian and the book is in portuguese, not sure if it has an english version. but the title is "Elementos de álgebra" by Arnaldo Garcia and Yves Lequain. I was gonna say if you want more information you can message me, but I don't think this site has this function, so I'll just leave it at that. – Rui Aldé Lopes Nov 08 '20 at 05:04

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