# Endomorphism image implies powering-invariant

Suppose $G$ is a group and $H$ is a subgroup of $G$ that is an endomorphism image of $G$, i.e., there exists an endomorphism $\sigma$ of $G$ such that $\sigma(G) = H$. Then, $H$ is a powering-invariant subgroup of $G$, i.e., if $G$ is powered over a prime $p$ (i.e., every element of $G$ has a unique $p^{th}$ root, so is $H$.