# Does $(\mathbb{Z} * \mathbb{Z} )/ \mathbb{Z} = (\mathbb{Z} \times \mathbb{Z})/\mathbb{Z} = \mathbb{Z}$?

My question is whether or not it's true that $(\mathbb{Z} * \mathbb{Z} )/ \mathbb{Z} = (\mathbb{Z} \times \mathbb{Z})/\mathbb{Z} = \mathbb{Z}$?

The problem I'm having in my head is trying to reconcile the following intuitions:

1. That it should be the case that $$(\mathbb{Z} * \mathbb{Z}) / \mathbb{Z} = \langle \, a, b \mid b=1 \rangle = \langle a \rangle = \mathbb{Z}$$
2. That the map \begin{align} \phi : (\mathbb{Z} \times \mathbb{Z}) / \mathbb{Z} & \to \mathbb{Z} \\ (a,b) & \mapsto a \end{align} should be a homomorphism such that $\ker(\phi) = \{ (0,b) | b \in \mathbb{Z} \} = \mathbb{Z}$
3. That $\mathbb{Z} * \mathbb{Z}$ is much bigger and more complex than $\mathbb{Z} \times \mathbb{Z}$ and taking the quotient of the same set should not result in the same group (I'm not sure if this is a general result)

Now, I'm not sure if I have been a bit too liberal in notation and should more concretely describe how $\mathbb{Z}$ is a subgroup in the free and direct products i.e. rewriting $(\mathbb{Z} * \mathbb{Z}) / \mathbb{Z}$ as $\langle a, b \rangle / \langle b \rangle$.

Either way I feel like i'm missing something fundamental how we take quotient groups and what those quotient groups tell us about the original group.

• $\mathbb{Z}$ isn’t a normal subgroup of the free product. – Qiaochu Yuan May 20 '18 at 4:30
• Oh of course, that answers the question, thankyou! – RichoKicked800goals May 20 '18 at 4:31
• Is it true that if we let $N$ be the smallest normal subgroup containing $\mathbb{Z}$ then $(\mathbb{Z} * \mathbb{Z} )/ N = \mathbb{Z}$? – RichoKicked800goals May 20 '18 at 4:34
• Yes, that’s right. – Qiaochu Yuan May 20 '18 at 4:44