Let $|G|$ be infinite. Use normal cores (not Poincaré's Theorem) to prove that if $H, K$ have finite index in $G$, then so does $H\cap K$.

Question

This is Exercise 7.4 of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to Approach0, it is new to MSE. This Google search yields this question, which is close but does not have the restriction to normal cores nor the prohibition of Poincaré's Theorem.

The Details:

On page 62 of Roman's book, there is

Poincaré's Theorem: Let $H_1, \dots, H_n\le G$ and $[G: H_i]<\infty$ for all $i$ and let $I_k=H_1\cap\dots\cap H_k$. Then Poincaré's Inequality holds:

$$[G: H_1\cap \dots\cap H_n]\le [G: H_1]\dots[G: H_n]$$

and so, in particular, $[G: H_1\cap \dots\cap H_n]$ is also finite.

1. The inequality above can be replaced can be replaced by division if $$I_kH_k\le G\tag{1}$$ for all $k\in\{1,\dots, n-1\}$.

2. Equality holds in Poincaré's Inequality if and only if $$[I_kH_{k+1}:H_{k+1}]=[G:H_{k+1}]$$ for all $k\in\{1,\dots, n-1\}$. Hence, if $G$ is finite or $(1)$ holds for all $k$, then equality holds in Poincaré's Inequality if and only if $$I_kH_{k+1}=G$$ for all $k\in\{1,\dots, n-1\}$.

On page 125 ibid., we have, within a theorem, the

Definition: Let $H\le G$. The largest subgroup of $H$ that is normal in $G$ is called the normal interior or core of $H$, which we denote by $H^\circ$.

The Question:

This question is from the chapter on group actions and the structure of $p$-groups, although I think the latter topic is not relevant here.

Exercise 7.4: Let $G$ be an infinite group. Use normal interiors (not Poincaré's Theorem) to prove that if $H$ and $K$ have finite index in $G$, then so does $H\cap K$.

Context:

As I said above, I think group actions are relevant here. Indeed, the theorem the definition above is in goes like this:

Theorem 4.19: Let $H\le G$. [. . .] The core of $H$ is the kernel $$H^\circ=\bigcap_{a\in G}H^a$$ of the action $\sigma$ of $G$ on $G/H$ by left translation.

Proof (p.125): We have

$$\begin{align} \ker(\sigma)&=\{x\in G\mid \sigma_x={\rm id}\}\\ &=\{x\in G\mid xaH=aH, \forall a\in G\}\\ &=\{x\in G\mid xa\in aH, \forall a\in G\}\\ &=\{x\in G\mid x\in a^{-1}Ha, \forall a\in G\}\\ &=\bigcap_{a\in G} H^a. \end{align}$$ $\square$

Thus, if I compose/compare somehow the actions $\sigma_L$ of $G$ on $G/L$ for $L\in\{H,K\}$ in question if possible, then I might get somewhere. I'm not sure how to flesh out this vague idea though.

Do I need to calculate $$(H\cap K)^\circ=\bigcap_{g\in G}(H\cap K)^g?$$ If so, I'm not sure how it helps.

(Here $L^g$ is $g^{-1}Lg$.)

Please help :)

Answer 1

Since $G/H$ is finite, $G/H^\circ$ is isomorphic to a subgroup of a finite symmetric group, i.e., $H^\circ$ is also of finite index. Likewise, $K^\circ$ is of finite index. Then the kernel of the obvious homomorphism $G\to G/H^\circ \times G/K^\circ$ is also of finite index and is clearly contained in $H\cap K$.