Normal of finite index implies completely divisibility-closed

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., normal subgroup of finite index) must also satisfy the second subgroup property (i.e., completely divisibility-closed normal subgroup)
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Statement

Suppose $G$ is a group and $H$ is a normal subgroup of finite index in $G$ , i.e., $H$ is a normal subgroup of $G$ and the index of $H$ in $G$ is finite. Then, $H$ is a completely divisibility-closed subgroup, and hence, a completely divisibility-closed normal subgroup of $G$ .

In other words, if $G$ is divisible by a prime number $p$ , then the quotient group $G / H$ is $p$ -torsion-free.

Related facts

Normal of finite index implies quotient-powering-invariant

Facts used

Divisibility is inherited by quotient groups
Finite and p-divisible implies p-powered (and hence also implies $p$ -torsion-free).

Proof

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Given: A group $G$ , a normal subgroup $H$ of $G$ such that the index of $H$ in $G$ is finite (in other words, the quotient group $G / H$ is a finite group). $G$ is divisible by a prime $p$ .

To prove: $G / H$ is $p$ -torsion-free.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$G / H$ is $p$ -divisible.	Fact (1)	$H$ is normal in $G$ , $G$ is $p$ -divisible.		Given-fact combination direct.
2	$G / H$ is $p$ -torsion-free.	Fact (2)	$H$ is normal of finite index, so $G / H$ is a finite group.	Step (1)	Step-fact-given direct.