Subgroup of Finite Index Contains a Normal Subgroup of Finite Index

Problem 232

Let $G$ be a group and let $H$ be a subgroup of finite index. Then show that there exists a normal subgroup $N$ of $G$ such that $N$ is of finite index in $G$ and $N\subset H$.

The group $G$ acts on the set of left cosets $G/H$ by left multiplication.
Hence it induces the permutation representation $\rho: G \to S_n$, where $n=|G:H|$.
(Note that a permutation representation is a group homomorphism.)
Let $N=\ker \rho$ be the kernel of the homomorphism $\rho$. Then $N \triangleleft G$.

By the first isomorphism theorem, the quotient group $G/N$ is isomorphic to a subgroup of $S_n$. In particular, $G/N$ is a finite group, hence the index $[G:N]$ is finite.

Finally, we show that $N \subset H$.
For any $x \in N=\ker \rho$, we have $x(gH)=gH$ for any $g \in G$.
In particular we have $xH=H$, hence $x \in H$.

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Hint.
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Proof.
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Proof.
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