TITLE: Why the group $\langle x,y\mid x^2=y^2\rangle $ is not free?
QUESTION [16 upvotes]: Why is the group $G= \langle x,y\mid x^2=y^2\rangle $ not free?

I can't find any reason like an element of finite order or some subgroup of it that is not free etc.

REPLY [3 votes]: This proof is "algebraically" deriving the statement "$G$ contains $\mathbb Z^2$, hence is not free" that Seirios's answer made.
First note that $x^2$ is in the center of $G$, and that $x^2$ is not trivial since we can consider a map $$G \to \mathbb Z, \quad x,y \mapsto 1,$$ in fact $x^2$ generates a group isomorphic to $\Bbb Z$ Consider the reflections $$a,b: \Bbb Z \to \Bbb Z, \quad a: t \mapsto -t, b: t \mapsto 2-t,$$ and consider the group $D_\infty = \langle a, b \rangle$, under function composition (this group happens to be isomorphic to $\Bbb Z/2 \Bbb Z* \Bbb Z/2 \Bbb Z$). Note that $ba: t \mapsto t+2$, so in particular, $ba$ generates a group isomorphic to $\mathbb Z$. So $yx$ generates a group isomorphic to $\mathbb Z$, if we consider $x\mapsto a, y \mapsto b$. Also note that no power of $yx$ is $x^2$, since $x^2$ is not the identity, but $x^2$ is in the kernel of the map, which no non-trivial power of $yx$ is in the kernel of the map. So we have that $\langle x^2, yx\rangle$ is an abelian group, with no torsion, generated by two elements, hence it is isomorphic to $\Bbb Z ^2$. Free groups do not contain subgroups isomorphic to $\Bbb Z^2$, and more generally hyperbolic groups do not contain subgroups isomorphic to $\Bbb Z^2$, hence $G$ is not free.
For a proof that $\Bbb Z^2$ can not be a subgroup of a hyperbolic group, look at proposition 3.5 in Notes on word hyperbolic groups by Alonso, Brady, Cooper, Ferlini, Lustig, Mihalik, Shapiro, and Short.