I am very bad with problems involving divisibility of orders and such. If anyone can give me some help with the following problem, it will be very much appreciated:
Prove that every subroup $H$ of $G$ of index $n$ must contain a normal subgroup $N \unlhd G$ such that $[G : N]$ divides $n!$ (this is a factorial and not an exclamation mark, just in case anyone gets confused as I did).
Answer
With $H$ come $n$ cosets of the form $gH$.
The group $G$ operates on the set of these cosets by left multiplication.
The kernel of this operation is a normal subgroup $N$ of $G$.
We have $N
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