# Show that if a group $G$ has odd order, any subgroup $H$ of index 3 in $G$ is normal in $G$? [duplicate]

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 $G$.

• 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
• Regarding the question posed in the edit: $|G| = |H|[G:H]$, so $3$ must divide $|G|$. – Bungo 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

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