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I have to show that $G=\langle x,y,z\ |xz=zx \rangle$ is not free. Now either I show it has an element of finite order which I dont see works here or I show it has no non-trivial defining relators i.e. I will need to prove that no Tietze transformation can reduce it to such group with no non-defining relators , but I cannot work it out.
But is there any other approach like we can use abelianness of $x$ and $z$.

$\textbf{My approach-} $ Is it correct? It clearly has $\mathbb{Z \oplus\ Z}$ as its subgroup, and every subgroup of free subgroup is free but $\langle x,z\ | xz=zx \rangle$ is not free as abelian and not cyclic.

Another problem is to show that $G= \langle x,y\ | x^2=y^2 \rangle$ is not free. Any hints for this one.

Bhaskar Vashishth
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1 Answers1

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The approach in the question works for the group $\langle x,z \mid xz=zx \rangle$, and to show that $G=\langle x,y \mid x^2 =y^2 \rangle$ is not free see this question that was later asked.

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