# Divisible abelian subgroup of abelian group contains no proper nontrivial verbal subgroup

## Statement

Suppose $G$ is an abelian group, and $H$ is a subgroup of $G$ that is a Divisible abelian group (?). Then, if $K$ is a subgroup of $H$ that is a verbal subgroup of $G$, then either $K$ is trivial or $K = H$.

## Facts used

1. verbal subgroup equals power subgroup in abelian group

## Proof

Given: An abelian group $G$, a subgroup $H$ of $G$ that is divisible, and a subgroup $K$ of $H$ that is verbal in $G$.

To prove: $K = H$ or $K$ is trivial.

Proof: By fact (1), there exists an integer $n$ such that $K$ is the set of $n^{th}$ powers in $G$. If $n = 0$, $K$ is trivial. If, on the other hand, $n$ is nonzero, then every element of $H$ is a $n^{th}$ power, so $H \le K$. Since $K \le H$ by assumption, we get $H = K$.