Given H is a subgroup of G and that H has finite index n, prove that there is a normal subgroup K of G such that $\displaystyle K \leq H$ and $\displaystyle |G/K| \leq n!$. Don't assume that G is finite.

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- Feb 18th 2011, 07:34 PMabhishekkgpnormal subgroups and cayley's theorem
Given H is a subgroup of G and that H has finite index n, prove that there is a normal subgroup K of G such that $\displaystyle K \leq H$ and $\displaystyle |G/K| \leq n!$. Don't assume that G is finite.

- Feb 18th 2011, 07:46 PMroninpro
Note that $\displaystyle G$ acts on the cosets of $\displaystyle H$ by left multiplication. Since there are $\displaystyle n$ cosets, this induces a homomorphism $\displaystyle \phi: G\to S_n$. Then we know that $\displaystyle K=\ker \phi$ is a normal subgroup of $\displaystyle G$.

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