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### H a subgroup of index n, then G has a normal subgroup K with [G : K] ≤ n!. [duplicate]

Prove that if G is a group and H a subgroup of index n, then G has a normal subgroup K with [G : K] ≤ n! I'm having trouble proving this because frankly I have no idea where to start. Any tips?
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### Show there is a normal subgroup $K$ of $G$ with $K\subset H$ and such that the order of $K$ divides $n!$ [duplicate]

Let $G$ be a finite group and suppose $H$ is a subgroup of $G$ having index $n$. Show there is a normal subgroup $K$ of $G$ with $K\subset H$ and such that the order of $K$ divides $n!$ . any ...
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### If $(G : H) = r$ and there exists $K \triangleleft G$ contained in $H$ such that $(G : K) = r!$ [duplicate]

Possible Duplicate: How to prove that if $G$ is a group with a subgroup $H$ of index $n$, then $G$ has a normal subgroup $K\subset H$ whose index in $G$ divides $n!$ I'm trying to prove the ...
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### Quotient group with normal subgroup dividing the order of another group [duplicate]

Let G be a group with subgroup H and let $\Omega$ be the set of right cosets of H in G. Show that if G is a group with a subgroup of index n then G has a normal subgroup with index dividing n! ...
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### Normal subgroup of index a divisor of n! [duplicate]

I’ve found this exercise on “basic algebra”: Show that if a finite group $G$ has a subgroup $H$ of index $n$ then $H$ contains a normal subgroup of $G$ of index a divisor of n!. My attempt: I’ve ...
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### Given a subgroup $H$, find a normal subgroup of $G$ with index less than or equal to $n!$ [duplicate]

Let $G$ be a group and $H$ a subgroup with finite index $n$. Prove that $G$ has a normal subgroup $N$ such that $N\subseteq H$and $|G: N| \le n!$.
### Prove that there is a subgroup $K$ : $k! | [H:K]$ [duplicate]
Consider a group $G$, and $H \in G$ - subgroup with $[G:H] = k$, then prove that there is exists normal subgroup $K$ in $H$ such that $k! | [G:K]$? Actually I have no ideas. Any hints?