Normal of finite index implies quotient-powering-invariant

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 quotient-powering-invariant subgroup of $$G$$, i.e., if $$G$$ is powered over a prime number $$p$$, then so is the quotient group $$G/H$$.

Related facts

 * Finite normal implies quotient-powering-invariant
 * Powering-invariant and central implies quotient-powering-invariant
 * Normal of finite index implies completely divisibility-closed

Facts used

 * 1) uses::Divisibility is inherited by quotient groups
 * 2) uses::Finite and p-divisible implies p-powered

Proof using given facts (less explicit)
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 powered over a prime $$p$$.

To prove: $$G/H$$ is also powered over $$p$$.

Proof:

Hands-on proof
This is the same as the preceding proof, but shows all the details explicitly without appeal to separate facts.

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). A prime number $$p$$ such that for any $$g \in G$$, there is a unique $$x \in G$$ such that $$x^p = g$$.

To prove: For every $$a \in G/H$$, there is a unique $$b \in G/H$$ such that $$b^p = a$$.

Proof: Let $$\varphi:G \to G/H$$ be the quotient map.