Show that $(\mathbb{Z}*\mathbb{Z})/[\mathbb{Z}*\mathbb{Z},\mathbb{Z}*\mathbb{Z}]\cong \mathbb{Z}\oplus\mathbb{Z}$ Im trying to show that
$$(\mathbb{Z}*\mathbb{Z)}/[\mathbb{Z}*\mathbb{Z},\mathbb{Z}*\mathbb{Z}]\cong \mathbb{Z}\times\mathbb{Z}$$
and my first thought was to use the first isomorphism theorem. I've been trying to use the homomorphism
$$\phi:\mathbb{Z}*\mathbb{Z}\to \mathbb{Z}\times \mathbb{Z}$$ defined by
$$\phi(a^{n_1}b^{m_1}\cdots a^{n_k}b^{m_k})=(a^{n_1}\cdots a^{n_k},b^{m_1}\cdots b^{m_k})$$ Where we are considering $\mathbb{Z}*\mathbb{Z}$ to be the free product of two infinite cyclic groups generated by a and b respectively. I have shown that this is a homomorphism and is surjective, but i am really struggling to show that the commutator is the kernel. I have that the commutator is contained in the kernel but i cannot show the other containment. I am beginning to suspect that it is in fact not equal to the kernel, but if that is the case I am stumped on how to proceed. I was hoping someone would be able to shine some light on the problem.
 A: Let $F_2 = \mathbb{Z} * \mathbb{Z}$, and let $F_2^{ab} = F_2/[F_2, F_2]$ denote its abelianization. For any abelian group $A$, we have
\begin{equation*}
\operatorname{Hom}(F_2^{ab}, A) = \operatorname{Hom}(F_2, A) = A\oplus A,
\end{equation*}
since $F_2$ is a free group of rank $2$. But $F_2^{ab}$ is a finitely-generated abelian group and thus must be of the form $\mathbb{Z}^n \oplus A_0$ for some integer $n\geq 0$ and some finite abelian group $A_0$, and it follows quickly that $F_2^{ab}$ must be $\mathbb{Z}^2$.
A: I ended up figuring out something using the first isomorphism theorem like I initially wanted to. Using the same homomorphism defined in my question, I was able to show that $[\mathbb{Z}*\mathbb{Z},\mathbb{Z}*\mathbb{Z}]=\ker(\phi)$. I did this using a lemma:
An element in $w\in\mathbb{Z}*\mathbb{Z}$, where
$$w=a^{n_1}b^{m_1}\cdots a^{n_k}b^{m_k}$$ is also an element of $[\mathbb{Z}*\mathbb{Z},\mathbb{Z}*\mathbb{Z}]$ if and only if
$$n_1+n_2+\cdots+n_k=0\,\,\,\text{and}\,\,\,m_1+m_2+\cdots+m_k=0$$
This lemma was fairly easy to prove using induction on the lengths of words where, to apply the induction step, you must multiply by some commutators to reduce the word length. Then with this tool it is easy enough to show that given some arbitrary $x\in \ker(\phi)$, we must have $x$ satisfying the right hand side of that iff and thus $\ker(\phi)=[\mathbb{Z}*\mathbb{Z},\mathbb{Z}*\mathbb{Z}]$. Then the first isomorphism theorem applies and yields the final result.
