## 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 $[G : K] \leq n!$.

• 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?

• nbloomf  On November 18, 2014 at 9:16 pm

Yes.