Finite normal implies quotient-powering-invariant

Statement
Suppose $$G$$ is a group and $$H$$ is a finite normal subgroup of $$G$$. Then, $$H$$ is a quotient-powering-invariant subgroup of $$G$$, i.e., for any prime number $$p$$ such that $$G$$ is powered over $$p$$, so is the quotient group $$G/H$$.

Facts used

 * 1) uses::Left cosets are in bijection via left multiplication

Proof idea
This is relatively straightforward, and involves using the fact that an injective map between subsets of equal cardinality (here, the subsets are cosets of $$H$$, and the map is the $$p^{th}$$ power map) must be surjective, and hence the inverse to it is defined. Note that the key way the proof fails for infinite group is that it is possible for the powering map between cosets of an infinite normal subgroup to be such that a given coset is a disjoint union of the images under the $$p^{th}$$ power map of more than one coset, even though all the maps are injective.

For instance, for the case $$G = \mathbb{Q}, H = \mathbb{Z}, p = 2$$, the identity coset is not covered as the set of squares of any single coset, but is the union of the set of squares of the cosets $$\mathbb{Z}$$ and $$\frac{1}{2} + \mathbb{Z}$$.

Proof details
Given: A group $$G$$, a finite normal subgroup $$H$$ of $$G$$. A prime number $$p$$ such that for any $$g \in G$$, there exists a unique $$x \in G$$ such that $$x^p = g$$.

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

Proof: