Subgroups of finite index force the existence of (possibly nontrivial) normal subgroups of bounded index

Prove that if a subgroup H has finite index n in a group G then there exists a normal subgroup K \vartriangleleft G such that K \leq H and [G : K] \leq n!.

G acts on the cosets G/H by left multiplication. Let \lambda : G \rightarrow S_{G/H} be the permutation representation induced by this action, and let K be the kernel of the representation.

Now K is normal in G, and K \leq \mathsf{stab}_G(H) = H. By the First Isomorphism Theorem, we have an injective group homomorphism \overline{\lambda} : G/K \rightarrow S_{G/H}. Since |S_{G/H}| = n!, we have [G : K] \leq n!.

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  • Jer  On December 8, 2010 at 4:05 pm

    Did you mean to write (gamma hat)=G/K -> S(G/H)?

    • nbloomf  On December 8, 2010 at 4:30 pm

      Yes. Thanks!

  • Gogo  On October 21, 2014 at 12:00 am

    I am sorry I am confused notation wise what is stab_G(H). Do you mean the stabilizer of H in G?

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