# Identifying $\langle x_1, x_2, x_3, x_4\mid x_1x_2=x_3, x_2x_1=x_3, x_1x_2=x_4, x_2x_1=x_4\rangle$.

What is the group $\langle x_1, x_2, x_3, x_4\mid x_1x_2=x_3, x_2x_1=x_3, x_1x_2=x_4, x_2x_1=x_4\rangle$?

It has infinite order according to GAP and the generators $x_3, x_4$ are redundant. Is it isomorphic to $\Bbb Z\times \Bbb Z$? It looks like it to me.

The relations imply that $x_1 x_2 = x_2 x_1$, so since $x_3, x_4$ are redundant the group is a quotient of $\{x_1, x_2 | x_1 x_2 = x_2 x_1\}$, which is the free abelian group $\Bbb Z \times \Bbb Z$ on two generators. On the other hand, declaring $x_3, x_4 := x_1 x_2$ lets one recover all of the given relations from $x_1 x_2 = x_2 x_1$, so, as you suggest, the group is exactly $\Bbb Z \times \Bbb Z$.