I am not sure that if there is a typo in Dummit's Abstract Algebra on page219 : Let $S=G$ and the map $\pi:F(S)\to G$ is the homomorphism extending the identity map of $S$ . the first paragragh writes " $R_0\leq\ker\pi$ " where $R_0$ is the set of words $g_ig_jg_k^{-1}$ , these $g$ are elements in the finite group $G$ satisfying $g_ig_j=_Gg_k$ .
It is clear that $R_0\subseteq\ker\pi$ , but how to show that $R_0\leq\ker\pi$ ? If so, we can also obtain $R_0\leq F(S)$ by $\ker\pi\leq F(S)$ , hence $R_0$ is also a free group since every subgroup of a free group is free. But I think " $R_0$ is the set of words $g_ig_jg_k^{-1}$ " restricts the length of each word in $R_0$ to be $\leq3$ , and this is a contradiction with that $R_0$ is free.