One example is, in fact, given in Exercise 3 (with the typo). Further, consider a free group $G$ with a basis $\{a_{1},\ldots,a_{n}\}$ . Take a new basis $\{2a_{1},a_{2},\ldots,a_{n}\}$ for a free subgroup $H$ of $G$ . If $x=2m_{1}a_{1}+m_{2}a_{2}+\cdots+m_{n}a_{n}$ , then $2m_{1},m_{2},\ldots,m_{n}$ are determined uniquely, and, hence, the equation has no more than one solution in $m_{1},\ldots,m_{n}$ . Further, if $a_{1}=2m_{1}a_{1}+m_{2}a_{2}+\cdots+m_{n}a_{n}$ , then $2m_{1}=1$ , and $m_{2}=\cdots=m_{n}=0$ , but there is no such integer $m_{1}$ , hence, $a_{1}\notin H$ , and $H$ is a proper free abelian subgroup of $G$ having rank $n$ .