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Let $H$ be a subgroup of $G $ with smallest possible prime index. Then $H$ is normal in $G$.

Above exercise is one of the classical exercises in group theory. The classical solution depends on group action on cosets of $H$. I want to share a proof not depending on group action. If you have other solutions, please share.

proof: If $H$ is not normal then assume $H\neq H^g$. Then $$|HH^g|=|H|\dfrac{|H|}{|H\cap H^g|}$$.

Notice that $\dfrac{|H|}{|H\cap H^g|}\geq p$. Thus, $|HH^g|\geq G$. We must have $HH^g=G$.

$g=hg^{-1}kg$ for some $h,k\in H$. Then $g=kh\in H$. As a result $H=G$ which is a contradiction.

mesel
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  • What is $H^g$? Do you mean $gHg^{-1}$? – D_S Apr 09 '16 at 17:56
  • @D_S: Yes, $H^g=g^{-1}Hg$ and to be able to say that $\dfrac{|H|}{|H\cap H^g|}\geq p$, I used the fact that $p$ is minimal. – mesel Apr 09 '16 at 17:58
  • At the end of the above, you get that $;g=kh\in H;$ ...but this was so only for this particular $;g\in G;$ for which $;H\neq H^g;$ . What about $;x\in H;$ if, for example, $;H=H^x;$ ? Say, for $;x\in N_G(H);$ ? – DonAntonio Apr 09 '16 at 18:33
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    The first part is similar to the way I did in my answer to http://math.stackexchange.com/questions/164244/normal-subgroup-of-prime-index/268484#268484 but you have found a nice way to avoid using that groups of order $p^2$ are abelian here. – Tobias Kildetoft Apr 09 '16 at 18:49
  • @Joanpemo The point is that the assumption was that for some $g\in G$ we had $H\neq H^g$ and this is what leads to the contradiction. – Tobias Kildetoft Apr 09 '16 at 18:49
  • Actually, the conclusion should be changed to $g\in H$ so $H = H^g$ which is the contradiction (you have indeed not shown that $H = G$). – Tobias Kildetoft Apr 09 '16 at 18:51
  • @TobiasKildetoft Thank you, exactly my point: we cannot deduce $;H=G;$ from the particular case of that particular $;g\in G;$ . – DonAntonio Apr 09 '16 at 20:37
  • @Joanpemo: when we see that $g\in H$, then $HH^g =HH=H=G$ which is also contradiction, I think ? – mesel Apr 09 '16 at 21:11
  • @mesel Again, no: you can not deduce $;H=G;$ because one particular $;g\in G;$ , namely: one for which $;H\neq H^g;$ is also in $;H;$ . Please read above and also Tobias' comments. – DonAntonio Apr 09 '16 at 21:39
  • @Joanpemo: Sometimes your proof can reach contradiction. You may not notice. But beauty of math is that you reached another contradicton. Again
    the above proof is not wrong.     – mesel Apr 09 '16 at 21:43
  • @Mesel Perhaps I'm missing something, but all I can see you can reach is, perhaps, that $;H=H^g;$ , which of course would also be a contradiction. – DonAntonio Apr 09 '16 at 21:45
  • @Joanpemo: By assuming $x$, you reached $y$ and by fact a, you can reach contradiction but you missed it. Then by $y$ you reached $z$ and by fact b you reached contradiction. I reach contradiction by the fact $H$ was a proper subgroup. (fact b) . You suggest that before that it contradict to fact that $H\neq H^g$. (fact a). Your suggestion is better yet what I mean is that above solution is not wrong from logical point of view. – mesel Apr 09 '16 at 22:44
  • I'm voting to close this question as off-topic because I had thought that it was beneficial for the people in this comunity yet I see that it is not very interesting for them beside that it is not a question. That is why I want to vote my post to be closed. – mesel Apr 10 '16 at 12:54
  • @mesel rather than voting to close your own question, you could just delete it? – Mathmo123 Apr 10 '16 at 13:36
  • You could add this as an answer to the question I linked. – Tobias Kildetoft Apr 10 '16 at 20:25
  • @TobiasKildetoft: Thanks, I added. – mesel Apr 12 '16 at 08:48

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