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Show that if a group $G$ has odd order, any subgroup $H$ of index 3 in $G$ is normal in $G$.

I think this is equivalent to the following: Let $H$ and $K$ be subgroups of a group $G$, with $K \leq H$. We can then show that $[G:K]=[G:H] \cdot [H:K]$. However, I'm not sure what the best approach is here. What's a good way to prove this theorem?

This question is different from Normal subgroup of prime index because we are not assuming that $3$ divides the order of $G$.

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user3749214
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    The only way I can see that the two statements are equivalent is in the fact that they are both true. For the one you are asking about, there is a general result that a subgroup of index $p$ where $p$ is the smallest prime dividing the order of the group will be normal (there are several ways to show that) – Tobias Kildetoft Apr 27 '16 at 19:57
  • I edited the question to explain why this is not a duplicate of "Normal subgroup of prime index". Can someone please explain why they think otherwise? – user3749214 Apr 27 '16 at 20:06
  • While it might not be an exact duplicate, it is answered by any answer to the linked question – Tobias Kildetoft Apr 27 '16 at 20:06
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    Regarding the question posed in the edit: $|G| = |H|[G:H]$, so $3$ must divide $|G|$. –  Apr 27 '16 at 20:07
  • @Bungo sorry but why does $|G|=|H|[G:H]$ imply that $3$ must divide $|G|$? – user3749214 Apr 27 '16 at 20:09
  • $[G:H] = 3$, no? –  Apr 27 '16 at 20:09
  • @Bungo ah yes, thank you. – user3749214 Apr 27 '16 at 20:11
  • I now see this is a duplicate of the linked question, so a moderator is welcome to close this question or whatever. Not sure exactly what the procedure is. Thanks everyone for the help! – user3749214 Apr 27 '16 at 20:12

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If the order of a subgroup is odd and the index of the subgroup is $3$ this implies that $3$ divides the order of the group.since $3$ is the least prime dividing the order of the group.hence every subgroup of index $3$ is normal

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