Let $U$ be a group with generators $\{a_1,\dots ,a_r\}$. If we want to show that $U$ is a free group, is it sufficent to show that $a_{i_1}^{k_1}a_{i_2}^{k_2}\dots a_{i_l}^{k_l} \neq 1$ where $k_i \in \mathbb{Z} - \{0\}$ and the successive $a_i$ are distinct? Sorry, I realize this could be more complete and technical statement, but I'm trying to remember a certain result...
Does anybody know what I'm thinking of?
Thanks!
This statement is equivalent to freeness of $U$ with free basis $a_1,...,a_n$. If you just want $U$ to be free and do not care about the basis then it is no longer true. For example $\Bbb{Z}$ is free of rank $1$ but it is generated by $2,3$ satisfying $2\cdot 2-3+2\cdot 2-3-2=0$.