One-relator group

Remarks

The relator may be taken cyclically reduced, since a cyclic permutation of $r$ is a conjugate of $r$ and generates the same normal subgroup. A one-relator group has an element of finite order other than $1$ if and only if $r$ is a proper power $s^n$ in $F(S)$ with $n \ge 2$. In that case, if $s$ is chosen not to be a proper power itself, then $s$ has order exactly $n$ in $G$.

Example: $\Z^2$ is a one-relator group. The map $\langle a, b \mid a b a^{-1} b^{-1} \rangle \to \Z^2$ with $a \mapsto (1, 0)$ and $b \mapsto (0, 1)$ is an isomorphism, since the relator says exactly that $a$ and $b$ commute.

Non-example: $\Z/2 \times \Z/2$ is not a one-relator group. If $\langle S \mid r \rangle$ with $S$ finite of size $k$ defined it, its abelianization would be $\Z^k$ modulo the subgroup generated by one vector, which is $\Z^{k-1} \oplus \Z/d$ for some integer $d \ge 0$. The group $\Z/2 \times \Z/2$ is abelian, so it equals its abelianization. It is finite, so $k - 1 = 0$, and then it would be cyclic, which it is not. A presentation with $S$ infinite gives an abelianization with infinitely many free factors $\Z$, so it fails as well.