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?
Subgroup of index 7
0
$\begingroup$
group-theory
finite-groups
-
1The 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? – 2012-01-03
-
1**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
1 Answers
2
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.
-
1In 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... – 2012-01-03
-
0could you tell me how can you solve the problem if we are allowed to take N=H, please? – 2012-01-03
-
0@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