$PROBLEM$ :If $G$ is finite of odd order. And $H$ subgroup of $ G$ with $[G:H]=3$ then $H$ is normal.
$Proof$:
I know that that there exists a homomorphism $\phi : G \rightarrow S_3 $ Such that $ker(\phi)<H$.I know that cause of i found it in the notes calling it Poincare's Trick.But i havent found a proof.(Id love to see a proof). Now that granted.
I know $\phi(G)<S_3$ and since $\phi$ not an isomorphism since $|G|=odd$ and $|S_3|=even$. So $|\phi(G)|\mid |S_3|$ so $|\phi(G)|=2$ or $3$.
Now i know $$G/ker\phi \simeq \phi(G) $$ meaning $|G/ker\phi |= |\phi(G)| $ And that must be an $odd$ since it is the $index$ of $ker\phi $ in $G$. So $$|\phi(G)|=|G/ker\phi|=3$$
Now i know $$|G|=[G:H]|H|$$ and $$|G|=[G:Ker\phi]|Ker\phi|$$ substituting with $3$. I get $$|H|=|Ker\phi| $$ and since $Ker\phi < H$ $\Rightarrow$ $Ker\phi=H$ .And i know $Ker\phi \lhd G$ So its $H$ too.
Am i wrong?Afraid of some sort of circular reasoning.Also id love to see a proof of why there is such homomorphism.
Im asking if my proof is correct and it is not a duplicate of the same problem.Since Im not asking for a proof but for a proof verification and a proof of a Trick mentioned!!
