# 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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"I have to find" suggests this is a homework problem or assignment; if this is the case, please consider using the [homework] tag. If it is not homework, then context would be helpful, given that you "have to find" such a group. – Arturo Magidin Jan 3 '12 at 2:04
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
@ArturoMagidin: it's not homework, I'm simply doing some exercises by myself to prepare better for an exam. – Alex M Jan 3 '12 at 2:21
@AlexM: Please provide this context, especially when you preface your post with words such as "I have to find", which definitely suggests assignments. You may also want to provide context in the form of the source of the problem, so that the question may be cast at an appropriate level. – Arturo Magidin Jan 3 '12 at 2:23

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