Central implies normal satisfying the subgroup-to-quotient powering-invariance implication

Original formulation
Suppose $$G$$ is a group and $$H$$ is a central subgroup of $$G$$. Note that central implies normal, so $$H$$ is normal. Then, $$H$$ is a normal subgroup satisfying the subgroup-to-quotient powering-invariance implication in $$G$$: if $$p$$ is a prime number such that both $$G$$ and $$H$$ are $$p$$-powered, so is the quotient group $$G/H$$.

Corollary formulation
Suppose $$G$$ is a group and $$H$$ is a subgroup of $$G$$ satisfying the following two conditions:


 * 1) $$H$$ is a central subgroup of $$G$$, i.e., every element of $$H$$ commutes with every element of $$G$$.
 * 2) $$H$$ is a powering-invariant subgroup of $$G$$, i.e., for any prime $$p$$ such that $$G$$ is powered over $$p$$, so is $$H$$.

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

Applications

 * Center is quotient-powering-invariant

Proof for original formulation
Given: A group $$G$$. A central subgroup $$H$$ of $$G$$A prime number $$p$$ such that $$G$$ is powered over $$p$$, i.e., every element of $$G$$ has a unique $$p^{th}$$ root in $$G$$. $$H$$ is also powered over $$p$$. Let $$\varphi:G \to G/H$$ be the quotient map. An element $$a \in G/H$$.

To prove: There is a unique element $$b \in G/H$$ satisfying $$b^p = a$$.

Proof:

Proof for corollary formulation
This is immediate from the original formulation, and the observation that:

powering-invariant subgroup and normal subgroup satisfying the subgroup-to-quotient powering-invariance implication $$\implies$$ quotient-powering-invariant subgroup

(actually, the converse implication also holds, but that is not necessary for us here).