TITLE: Showing $H\unlhd G$ when $[G\colon H]$ is not the smallest prime dividing $|G|$
QUESTION [9 upvotes]: I recently read about the theorem that for a finite group $G$, if $p$ is the least prime dividing $|G|$, then any subgroup $H$ with $[G\colon H]=p$ is normal in $G$.
Going over some exercises, this theorem made it easy to see that a subgroup $H$, with $|H|=35$, is normal in $G$ when $|G|=70$. 
A similar situation wasn't as obvious. If $|H|=35$, but $|G|=140$, then $H$ has index $4$. How then would one go about proving $H\unlhd G$? Thanks.

REPLY [2 votes]: I think I may have thought up an answer which doesn't use a whole lot of group theory. If anyone reads this, I'd appreciate feedback on whether it's right or wrong.
Let $H$ act on $G/H$ by $h\cdot gH=hgH$. Since $[G:H]=4$, the orbit of any $gH\in G/H$ is at most $4$. But the order of any orbit must divide $|H|$, so must be $1$, $5$, $7$, or $35$. So each orbit is of order 1, and thus trivial since $egH=gH$ in particular. So $hgH=gH$ for any $g\in G$, $h\in H$, hence $g^{-1}hgH=g^{-1}gH=H$. Then $g^{-1}hg\in H$, so $H\unlhd G$?