# Divisibility is quotient-closed

Suppose $G$ is a group and $H$ is a normal subgroup of $G$, so that $G/H$ is the quotient group. Then, if $n$ is a natural number such that $G$ is $n$-divisible (i.e., every element of $G$ has a $n^{th}$ root in $G$, then so is $G/H$, i.e., every element of $G/H$ has a $n^{th}$ root in $H$.
Given: A group $G$ and a natural number $n$ such that every element of $G$ has a $n^{th}$ root in $G$, a normal subgroup $H$ with quotient group $G/H$. An element $a \in G/H$.
To prove: There exists $b \in G/H$ such that $b^n = a$.
Proof: Let $\varphi:G \to G/H$ be the quotient map and let $g \in G$ be such that $\varphi(g) = a$. Since $G$ is $n$-divisible, there exists $h \in G$ such that $h^n = g$. Let $b = \varphi(h)$. Then, $b^n = (\varphi(h))^n = \varphi(h^n) = \varphi(g) = a$, so $b$ is as desired.