Normal of finite index implies completely divisibility-closed

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

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

Proof
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: