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This question already has an answer here:

Assume that $G$ is finite with $p$ the smallest prime dividing its order. Suppose $H < G$ with $[G:H]=p$. Prove that $H \lhd G$.

I've seen this question a few times on here but all the proofs I saw appeal to the First Isomorphism Theorem. I had old class notes showing this proved solely with group actions and orbits but I could only make out some of it so I recreated it from scratch. Can you guys check this for me?

Let $H$ act on the set $G/H$ by left multiplication. Consider the orbit of $H$ itself. $\mathcal{O}_H=\{hH \mid h\in H\}$ and $hH = H$, thus $\mathcal{O}_H$ contains one coset, namely, $H$. Let $x\in G - H$. What can be said about $\mathcal{O}_{xH}$? This action of left multiplication partitions $G/H$ into disjoint orbits of cosets. Because $|G/H| = p$, the sum of the sizes of the orbits is equal to $p$. We already know that $|\mathcal{O}_H| = 1$ therefore $|\mathcal{O}_{xH}| \leq p-1$. The Orbit-Stabilizer theorem gives us $|\mathcal{O}_{xH}| |Stab_{xH}| = |H|$. We see that $|\mathcal{O}_{xH}|$ divides $|H|$ and hence divides $|G|$. But since $p$ was the smallest prime dividing the order of $G$, $|\mathcal{O}_{xH}| = 1$.

The immediate consequence is that $|Stab_{xH}| = |H|$. As $Stab_{xH} \leq H$, we have $Stab_{xH} = H$. Therefore for every $h \in H$, $hxH=xH$. This implies $x^{-1}hx\in H$ and thus $x^{-1}Hx = H$ for all $x \in G$. Therefore, $H \lhd G$.

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marked as duplicate by Derek Holt group-theory Aug 15 '17 at 15:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ math.stackexchange.com/q/164244/8581 $\endgroup$ – mrs Jun 14 '16 at 4:01
  • $\begingroup$ math.stackexchange.com/a/289507/8581 $\endgroup$ – mrs Jun 14 '16 at 4:02
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    $\begingroup$ Seeing as the Orbit-Stabilizer theorem is proved using the first isomorphism theorem, I don't really see how this gets around the core issue stated at the beginning of the post. $\endgroup$ – Alex Wertheim Jun 14 '16 at 4:05
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    $\begingroup$ I want to avoid outright appealing to a homomorphism to the symmetric group. I hear what you mean though. $\endgroup$ – Bo Rel Jun 14 '16 at 17:57
  • $\begingroup$ This seems okay to me $\endgroup$ – Sushil Feb 16 '18 at 18:42
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easily you can prove that four predicate in blow are equivalent .1-H is normal in G .2-for every element of G example a we have a^p is element of H .3-for every element of G for example a exist a natural number n(a) that (a)^n(a) is element of H that if q|n(a)(q is a prime number) then q>=p. 4- for every element of G for example a we have a,a^2,a^3,...,a^p-1 is'nt element of H. you can prove that 1»»2 and 2»»3 and 3»»4 and 4»»1

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  • $\begingroup$ Right, but the OP wanted you to check the way! $\endgroup$ – mrs Jun 14 '16 at 4:23
  • $\begingroup$ I write the my solution completly? $\endgroup$ – habib alizadeh Jun 14 '16 at 4:28
  • $\begingroup$ babak you are iranian? $\endgroup$ – habib alizadeh Jun 14 '16 at 4:30
  • $\begingroup$ Thanks for the post, but I'm not looking for alternative proofs. I just wanted someone to read through my proof and check the logic. $\endgroup$ – Bo Rel Jun 15 '16 at 1:58
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There is a much simpler way to prove this:

Proposition Let $p$ be the smallest prime dividing the order of the finite group $G$, and assume that the subgroup $H$ has index $p$. Then $H \lhd G$.

Lemma Let $H \leq G$ with $G=HH^g$ for some $g \in G$. Then $G=H$.

Proof Apparently $g=hk^g$, for some $h,k \in H$. Hence $g=hg^{-1}kg$, from which one derives that $g=kh \in H$. Hence $H^g=H$, so $G=HH^g=HH=H$.

Let us proceed with a proof of the Proposition: assume that $H$ is not normal, so there exists a $g \in G$ with $H^g \ne H$. Since $H \ne G$, we have $|G| \gt |HH^g|$. But $|HH^g|=\frac{|H| \cdot |H^g|}{|H \cap H^g|}$. This shows that $|G:H|=p \gt |H^g:H \cap H^g|$. We conclude that $|H^g:H \cap H^g|=1$, as $p$ is the smallest prime dividing $|G|$. This means that $H^g=H \cap H^g$, so $H^g \subseteq H$, implying $H=H^g$, since $H^g$ and $H$ have the same order. We arrive at a contradiction.

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