Complemented normal implies quotient-powering-invariant

Statement

Suppose $G$ is a group and $H$ is a complemented normal subgroup of $G$ (i.e., there exists a permutable complement to $H$ in $G$ , i.e., $G$ is an internal semidirect product involving $H$ ). Then, $H$ is a quotient-powering-invariant subgroup of $G$ : for any prime number $p$ such that $G$ is powered over $p$ (i.e., every element of $G$ has a unique $p^{th}$ root), $G/H$ is also powered over $p$ .

Facts used

Proof

Proof using given facts

The proof follows directly from Facts (1) and (2).

Hands-on proof

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Given: A group $G$ , a normal subgroup $H$ of $G$ with permutable complement $K$ . $G$ is powered over a prime number $p$ .

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

Proof:

Step no.	Assertion/construction	Given data used	Previous steps used	Explanation
1	Let $\varphi :G\to K$ be the retraction that sends an element of $G$ to the unique element of $K$ in its coset with respect to $H$ . This corresponds to the quotient map $G\to G/H$ and $G/H\cong K$ .	$H$ is normal in $G$ and is complemented, with complement $K$ .
2	For $g\in K$ , there exists a unique $x\in G$ such that $x^{p}=g$ .	$G$ is powered over $p$ .
3	The $x$ obtained in Step (2) is actually an element of $K$ . In other words, for $g\in K$ , there exists a unique $x\in K$ such that $x^{p}=g$ . So, $K$ is powered over $p$ .		Steps (1), (2)	We have that $\varphi (x)^{p}=\varphi (x^{p})=\varphi (g)=g$ . Thus, $\varphi (x)$ is an element whose $p^{th}$ power is $g$ . Since $x$ is the unique element of $G$ such that $x^{p}=g$ , we must have $\varphi (x)=x$ , forcing $x\in K$ .
4	$G/H$ is powered over $p$ .		Steps (1), (3)	Step (1) says that $G/H\cong K$ . Step (3) says that $K$ is powered over $p$ . Thus, $G/H$ is powered over $p$ .