# 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 $p$.

## 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 Facts used 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 $p$.