Powering-invariant over quotient-powering-invariant implies powering-invariant

Statement
Suppose $$G$$ is a group and $$H,K$$ are subgroups of $$G$$ such that $$H \le K \le G$$, $$H$$ is normal in $$G$$, and the following conditions hold:


 * $$H$$ is a quotient-powering-invariant subgroup of $$G$$.
 * $$K/H$$ is a powering-invariant subgroup of $$G/H$$.

Related facts

 * Powering-invariance is not quotient-transitive
 * Quotient-powering-invariance is quotient-transitive

Proof
Given: $$G$$ is a group and $$H,K$$ are subgroups of $$G$$ such that $$H \le K \le G$$, $$H$$ is normal in $$G$$, and the following conditions hold:


 * $$H$$ is a quotient-powering-invariant subgroup of $$G$$.
 * $$K/H$$ is a powering-invariant subgroup of $$G/H$$.

$$p$$ is a prime number such that $$G$$ is powered over $$p$$. An element $$g \in K$$.

To prove: There exists $$x \in K$$ such that $$x^p = g$$.

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