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If a group $G$ contains a subgroup ($\neq G$) of finite index, it contains a normal subgroup ($\neq G$) of finite index.

I have written a solution below. I used proposition 4.8 section 2.4 and portion of proof of corollary 4.10 section 2.4 to prove above exercise. Statement of proposition 4.8 is included in answer. My proof don’t reprove any results which have been proven in that section.

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  • $\begingroup$ See also these duplicates. $\endgroup$
    – Dietrich Burde
    Commented Sep 15, 2023 at 15:31
  • $\begingroup$ @DietrichBurde Thank you for the link. That post is indeed helpful. Certainly exercise in that post is same as of this post, solution intended by OP of that post is different from mine. Though I learned a new way to prove proposition 4.8. $\endgroup$
    – user264745
    Commented Sep 15, 2023 at 16:31
  • $\begingroup$ I think using the normal core is a natural way to prove this result. $\endgroup$
    – Dietrich Burde
    Commented Sep 15, 2023 at 16:57

1 Answer 1

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Proposition 4.8. Let $H$ be a subgroup of a group $G$ and let $G$ act on the set $S$ of all left cosets of $H$ in $G$ by left translation. Then the kernel of the induced homomorphism $\varphi:G\to A(S)$ is contained in $H$, where $A(S)$ is a group of all permutation of $S=\{gH\mid g\in G\}$.

Proof: Suppose $H(\neq G)$ be a subgroup of $G$ with $[G:H]=n\in \Bbb{N}$. Let $K=\ker \varphi$. By proposition 4.8, $K\subseteq H$. Since $[G:H]=n$, it is easy to check $A(S)\cong S_n$. By first isomorphism theorem, $G/K\cong \text{Im} \varphi \leq A(S)\cong S_n$. So $|G/K|=|\text{subgroup of }A(S)|$ divides $|S_n|=n!$. Which implies $|G/K|=[G:K]$ is finite, $1\leq [G:K]\leq n!$ to be precise. Clearly $K\neq G$, since $K\subseteq H\subset G$. Thus $\exists K(\neq G)$ subgroup of $G$ such that $K\subseteq H$, $K\lhd G$ and $[G;K]$ is finite.

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