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 $G/H$.

Then, $K$ is a powering-invariant subgroup of $G$.

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 $\phi :G\to G/H$ be the quotient map.

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	There exists $x\in G$ such that $x^p=g$.		$G$ is $p$-powered.		Given-direct.
2	$G/H$ is $p$-powered.		$H$ is quotient-powering-invariant in $G$, $G$ is $p$-powered.		Given-direct.
3	$(\phi (x){)}^p=\phi (g)$ and $\phi (x)$ is the unique $p^{th}$ root of $\phi (g)$ in $G/H$.			Steps (1), (2)	Applying the homomorphism $\phi$ to Step (1) gives that $(\phi (x){)}^p=\phi (g)$. Since, by Step (2), $G/H$ is $p$-powered, $\phi (x)$ must be the unique $p^{th}$ root.
4	$\phi (x)\in K/H$.		$K/H$ is powering-invariant in $G/H$.	Steps (2), (3)	By Step (2), $G/H$ is $p$-powered, hence $K/H$ is $p$-powered from the given information. Thus, the unique $p^{th}$ root of $\phi (g)$ in $G/H$, which equals $\phi (x)$,(from Step (3)) must be in $K/H$.
5	$x\in K$.		$H\le K$	Step (4)	Since $\phi (x)\in K/H$, $x\in {\phi }^{-1}(K/H)$, which is $K$ since $H\le K$.