Abstract Algebra – Why is the Group $\langle x,y\mid x^2=y^2\rangle$ Not Free?

Question

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.

Answer 1

The abelianization of a free group is a free abelian group: in particular, it is torsion-free. For your presentation, the relation $$2(x-y)=0$$ holds in the abelianization. Therefore, $G$ is free iff it is infinite cyclic. However, there clearly exists a morphism from $G$ onto $\mathbb{Z}_2 \times \mathbb{Z}_2$, so $G$ cannot be cyclic.

Added: In fact, your group $G$ is the fundamental group of the connected sum of two projective planes, that is of the Klein bottle $K$.

But we know that there exists a two-sheeted covering $\mathbb{T}^2 \to K$ from the torus $\mathbb{T}^2$. Therefore, $G$ has a subgroup of finite index two isomorphic to $\mathbb{Z}^2$. In particular, $G$ cannot be free.

Algebraically, it can be verified that $\langle x^2,xy^{-1} \rangle$ is a subgroup isomorphic to $\mathbb{Z}^2$ of index two in $\langle x,y \mid x^2=y^2 \rangle$.