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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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    **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 $[H:N]\geq 6!$?2012-01-03

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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.

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    @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.2012-01-03