Let $G$ be a finite group. Let $G_1$ be a subgroup of $G$ such that $[G:G_1]=2$. Now suppose that $H$ is a normal subgroup of $G_1$ with odd order. Prove that $H$ is a normal subgroup of $G$.
Some remarks/partial progress:
I have shown that if $G$ is a group, $G_1$ is a subgroup of index $2$, and $H$ is a subgroup of odd order, then $H\subset G_1$. I don't think this is relevant to the problem at hand though.
If $H$ is a group of odd order, squaring is injective.
Since $G_1$ in the original problem statement has index $2$, any element $g\in G$ satisfies $g^2 \in G_1$.
If $g\in G$ is in $G_1$, then clearly $gH=Hg$, so it suffices to show our claim for $g\in G\setminus G_1$. Note further than $G\setminus G_1$ is the set of all elements of even order.
The answer is "no". Take the wreath product $G=C_3\wr C_2$ of order $18$. It has a subgroup $G_1=C_3\times C_3$ of index $2$. The subgroup $G_1$ has several subgroups $H$ of order 3, each of which is normal in the Abelian group $G_1$ but only one of which is normal in $G$.