Showing $H\unlhd G$ when $[G\colon H]$ is not the smallest prime dividing $|G|$ 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.
 A: Since $[G:H]=4$, there is a homomorphism $\phi \colon G\rightarrow S_4$ with $ker(\phi)\subseteq H$. (See "Generalized Cayley's theorem- Introduction to theory of groups : Rotman )
If $ker(\phi)\neq H$, then it will be a subgroup (of $H$) of order $1,5$ or $7$. 


*

*If $|Ker(\phi)|=7$ then $|Im(\phi)|=|G/ker(\phi)|=140/7=20$; but $S_4$ has no subgroup of order 20.

*If $|Ker(\phi)|\leq 5$ then $|Im(\phi)|=|G/ker(\phi)|\geq 140/5>24$, contradiction.
Therefore $ker(\phi)=H$ which is normal in $G$.
A: A $7$-Sylow subgroup of $G$ has order $7$; the number of $7$-Sylow subgroups must be congruent to $1$ modulo $7$, and must divide $140 = 7\times 5\times 4$. Thus, it must divide $20$ and be congruent to $1$ modulo $7$; the only possibility is that the number of $7$-Sylow subgroups of $G$ is one (hence it is normal; in fact, characteristic) Thus, the subgroup $H$ of order $35$ must contain the unique $7$-Sylow subgroup of $G$.
Likewise, a $5$-Sylow subgroup of $G$ has order $5$, and the number of $5$-Sylow subgroups of $G$ must be congruent to $1$ modulo $5$ and divide $140$, hence must divide $7\times 4$. The only possibility is that there is a unique $5$-Sylow subgroup of $G$ (which is therefore characteristic, and thus normal), which must also be contained in $H$.
Thus, $H$ must be product of the unique $7$-Sylow subgroup and the unique $5$-Sylow subgroup of $G$. Since $H$ is the product of characteristic subgroups, it is in fact characteristic and therefore normal in $G$.
(Note that any group of order 35 is necessarily cyclic, by the by).
A: 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 $H\unlhd G$?
