[proofplan]
We use the defining quotient model $G^{\mathrm{ab}}=G/[G,G]$. Since $A$ is abelian, every commutator in $G$ is sent by $\varphi$ to the identity of $A$, so the whole commutator subgroup lies in $\ker \varphi$. This lets us define the induced map on cosets by $g[G,G]\mapsto \varphi(g)$; the kernel containment is exactly the condition that makes this definition independent of the representative. Finally, the quotient map $\pi$ is surjective, so any map factoring $\varphi$ through $\pi$ is forced to agree with this induced map on every coset.
[/proofplan]
[step:Show that every commutator lies in the kernel of $\varphi$]
Let $e_G$ denote the identity element of $G$, and let $e_A$ denote the identity element of $A$. Put
\begin{align*}
N := [G,G].
\end{align*}
Define the kernel subgroup of $\varphi$ by
\begin{align*}
K := \ker\varphi = \{g \in G : \varphi(g)=e_A\}.
\end{align*}
For $g,h \in G$, define the commutator of $g$ and $h$ by
\begin{align*}
[g,h] := g h g^{-1} h^{-1}.
\end{align*}
Since $\varphi: G \to A$ is a [group homomorphism](/page/Group%20Homomorphism) and $A$ is abelian, we have
\begin{align*}
\varphi([g,h]) = \varphi(g)\varphi(h)\varphi(g)^{-1}\varphi(h)^{-1}.
\end{align*}
Using commutativity in $A$, this becomes
\begin{align*}
\varphi([g,h]) = \varphi(g)\varphi(g)^{-1}\varphi(h)\varphi(h)^{-1} = e_A.
\end{align*}
Thus every commutator lies in $K$. Since $N=[G,G]$ is the subgroup generated by all commutators and $K \le G$, it follows that
\begin{align*}
N \subseteq K.
\end{align*}
[guided]
The purpose of this step is to identify the exact obstruction to factoring $\varphi$ through the quotient by $[G,G]$. Elements of $G^{\mathrm{ab}}=G/[G,G]$ forget commutators, so $\varphi$ can descend to this quotient only if $\varphi$ sends those commutators to the identity.
Let $e_G$ be the identity element of $G$, let $e_A$ be the identity element of $A$, and define
\begin{align*}
N := [G,G].
\end{align*}
Define the kernel subgroup of $\varphi$ by
\begin{align*}
K := \ker\varphi = \{g \in G : \varphi(g)=e_A\}.
\end{align*}
For $g,h \in G$, the commutator of $g$ and $h$ is
\begin{align*}
[g,h] := g h g^{-1} h^{-1}.
\end{align*}
Because $\varphi: G \to A$ is a group homomorphism, it respects products and inverses. Therefore
\begin{align*}
\varphi([g,h]) = \varphi(g h g^{-1} h^{-1}) = \varphi(g)\varphi(h)\varphi(g)^{-1}\varphi(h)^{-1}.
\end{align*}
Now we use the hypothesis that $A$ is abelian. The elements $\varphi(g)$ and $\varphi(h)$ commute in $A$, so we may move $\varphi(g)^{-1}$ next to $\varphi(g)$:
\begin{align*}
\varphi(g)\varphi(h)\varphi(g)^{-1}\varphi(h)^{-1} = \varphi(g)\varphi(g)^{-1}\varphi(h)\varphi(h)^{-1}.
\end{align*}
The inverse identities in $A$ give
\begin{align*}
\varphi([g,h]) = e_A.
\end{align*}
Thus every commutator belongs to $K$. Since $N=[G,G]$ is, by definition, the subgroup generated by all such commutators, and since $K$ is a subgroup of $G$, every product of commutators and inverse commutators also lies in $K$. Hence
\begin{align*}
N \subseteq K.
\end{align*}
[/guided]
[/step]
[step:Define the induced map on cosets and prove it is well-defined]
Define a function
\begin{align*}
\overline{\varphi}: G^{\mathrm{ab}} \to A, \qquad gN \mapsto \varphi(g).
\end{align*}
We verify that this definition is independent of the chosen representative. Suppose $g_1,g_2 \in G$ satisfy $g_1N=g_2N$. Then $g_2^{-1}g_1 \in N$. Since $N \subseteq K=\ker\varphi$, we have
\begin{align*}
\varphi(g_2^{-1}g_1)=e_A.
\end{align*}
Applying the homomorphism property,
\begin{align*}
\varphi(g_2)^{-1}\varphi(g_1)=e_A.
\end{align*}
Multiplying on the left by $\varphi(g_2)$ gives
\begin{align*}
\varphi(g_1)=\varphi(g_2).
\end{align*}
Therefore $\overline{\varphi}$ is well-defined.
[/step]
[step:Verify that $G^{\mathrm{ab}}$ is abelian and that the induced map factors $\varphi$]
Let $g,h \in G$. The product in $G^{\mathrm{ab}}=G/N$ is given by
\begin{align*}
(gN)(hN)=(gh)N.
\end{align*}
We first prove that $G^{\mathrm{ab}}$ is abelian. Since
\begin{align*}
g^{-1}h^{-1}gh=[g^{-1},h^{-1}],
\end{align*}
the element $g^{-1}h^{-1}gh$ lies in $N=[G,G]$. Also
\begin{align*}
hg(g^{-1}h^{-1}gh)=gh.
\end{align*}
Therefore
\begin{align*}
ghN=hg(g^{-1}h^{-1}gh)N=hgN.
\end{align*}
Equivalently,
\begin{align*}
(gN)(hN)=(hN)(gN).
\end{align*}
Since $gN$ and $hN$ were arbitrary elements of $G/N$, the group $G^{\mathrm{ab}}=G/N$ is abelian.
Using the definition of $\overline{\varphi}$ and the homomorphism property of $\varphi$, we obtain
\begin{align*}
\overline{\varphi}((gN)(hN))=\overline{\varphi}((gh)N)=\varphi(gh)=\varphi(g)\varphi(h)=\overline{\varphi}(gN)\overline{\varphi}(hN).
\end{align*}
Thus $\overline{\varphi}:G^{\mathrm{ab}}\to A$ is a group homomorphism. Since $G^{\mathrm{ab}}$ was proved abelian above and $A$ is abelian by hypothesis, this is a homomorphism of abelian groups.
For every $g\in G$,
\begin{align*}
(\overline{\varphi}\circ\pi)(g)=\overline{\varphi}(gN)=\varphi(g).
\end{align*}
Hence
\begin{align*}
\varphi=\overline{\varphi}\circ\pi.
\end{align*}
[/step]
[step:Use surjectivity of the quotient map to prove uniqueness]
Let $\psi_1:G^{\mathrm{ab}}\to A$ and $\psi_2:G^{\mathrm{ab}}\to A$ be homomorphisms of abelian groups such that
\begin{align*}
\psi_1\circ\pi=\varphi
\end{align*}
and
\begin{align*}
\psi_2\circ\pi=\varphi.
\end{align*}
Let $x\in G^{\mathrm{ab}}$. Since $\pi:G\to G^{\mathrm{ab}}$ is the quotient map, it is surjective, so there exists $g\in G$ such that
\begin{align*}
x=\pi(g).
\end{align*}
Then
\begin{align*}
\psi_1(x)=\psi_1(\pi(g))=(\psi_1\circ\pi)(g)=\varphi(g).
\end{align*}
Also,
\begin{align*}
\psi_2(x)=\psi_2(\pi(g))=(\psi_2\circ\pi)(g)=\varphi(g).
\end{align*}
Therefore $\psi_1(x)=\psi_2(x)$ for every $x\in G^{\mathrm{ab}}$, and hence $\psi_1=\psi_2$. This proves uniqueness and completes the proof.
[/step]
