# Subgroup of index 7

I have to find a group $G$ and a subgroup $H$ with $[G:H]=7$ such that for every $N$ normal subgroup of $G$ with $N\subset H$ we have $[G:N]\geq 7!$, could you help me please?

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The simplest way this could happen is if there are no nontrivial normal subgroups contained in $H$, and $G$ has order at least $7!$. Can you think of a group of order $7!$ with few normal subgroups? – Chris Eagle Jan 3 '12 at 2:04
Hint. If $N$ is normal in $G$ and contained in $H$, then it is also normal in $H$. Can you think of a group $H$ with the property that if $N\triangleleft H$, $N\neq H$, then $[H:N]\geq 6!$? – Arturo Magidin Jan 3 '12 at 2:05

Pick a simple group $S$ of order larger that $7!$ and let $G=C_7\times S$ be the direct product of $S$ and a cyclic group of order $7$. Pick $H$ to be $S\subseteq G$, which has index $7$ in $G$; there are no non-trivial normal subgroups of $G$ contained in $H$, so your second condition more or less vacuously holds.
In fact, you just need $|S|\geq 6!$. If you are allowed to take $N=H$ (unclear in the original post), then a semidirect product and a different $H$ will do... – Arturo Magidin Jan 3 '12 at 2:13
@Alex: Can you think of a group $G$ of order $7!$ with very few normal subgroups, that has a subgroup of $H$ index $7$, with $H$ having very few normal subgroups, no nontrivial one of which is normal in $G$? It's a very standard group. – Arturo Magidin Jan 3 '12 at 6:07