Group with relations, must identity elements be combination of the relations Let $G=\langle g_1,\dots, g_m\mid r_1,r_2,\dots,r_k\rangle$, where $r_i$ are relations, e.g. $r_1=g_{2}^{-1}g_3g_4=1$.
Suppose $g_{i_1}\cdots g_{i_n}=1$, that is, a product of generators that is equal to the identity.
Further suppose $g_{i_1}\cdots g_{i_n}$ is a reduced word, that is, there are no such terms like $gg^{-1}$ in the expression.
Can we conclude that $g_{i_1}\cdots g_{i_n}$ is a product of the relations $r_i$? Is this obvious or does it need a proof?
Thanks.
 A: No, you may not conclude that $g_{i_1}\cdots g_{i_n}$ is a product of relations $r_i$.
Instead, the following holds. I will write $U=_GW$ to mean the words $U$ and $V$ define the same element of the group $G$, while $U\equiv V$ means they define the same element of the ambient free group.

If $g_{i_1}\cdots g_{i_n}=_G1$ then there exist $h_1, \ldots, h_m\in F(g_j)$ such that $g_{i_1}\cdots g_{i_n}\equiv h_{j_1}r_{j_1}h_{j_1}^{-1}\cdots h_{j_p}r_{j_p}h_{j_p}^{-1}$.

That is, your word is the product of conjugates of relators, not simply the product of relators. This is because the presentation $G=\langle g_1, \ldots, g_m\mid r_1, \ldots, r_k\rangle$ is notation for the quotient group $F(g_1, \ldots, g_m)/N$ where $N$ is the normal closure of the elements $r_1, \ldots, r_k\in F(g_1, \ldots, g_m)$ (often written $N=\langle\langle r_1, \ldots, r_k\rangle\rangle$).
For more details, see Section 2.1 of the book Combinatorial group theory by Magnus, Karrass and Solitar. You should also read over Chapter 1 - it is gory, but everyone should read it once so you understand the pedantry surrounding group presentations.
