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Subgroups of finite index force the existence of normal subgroups of bounded index


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.2 Exercise 4.2.8

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


Linearity

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