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I'm trying to understand this statement. The free group $F_2=\langle a,b\rangle$ is freely generated, so by the definition there is a homomorphism from $F_2$ to $\mathbb{Z}\oplus\mathbb{Z}=\{c^m,d^n\mid m,n\in\mathbb{Z}\}$, sending $a$ to $c$ and $b$ to $d$. Let $K$ be its kernel and $N$ is the normal closure of $aba^{-1}b^{-1}$. Thus $$\mathbb{Z}\oplus\mathbb{Z}\cong F_2/K, \quad \langle a,b\mid aba^{-1}b^{-1}\rangle\cong F_2/N.$$ Since $N$ is the smallest normal subgroup that contains $aba^{-1}b^{-1}$, it is clear that $N\subset K$.

Now my question is how to verify $K\subset N$.

In Bowditch's note A course on geometric group theory says $F_2/N$ is abelian (I think it is because of $aba^{-1}b^{-1}$), and it is generated by $Na$ and $Nb$ which commute. Thus a typical element has the form $Na^mb^n$. This gets sent to $c^md^n$ under the natural map to $\mathbb{Z}\oplus\mathbb{Z}=F_2/K$. But I'm completely confused by the following statement:

If this is the identity, then $m=n=0$. This shows that $N=K$ as claimed.

Why do we consider the identity and how do we get the equation? Then why can we deduce the inclusion relationship from this equation?

itkyitfbku
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    $$ \boxed{;F_2/[F_2,F_2];\cong;\mathbb Z\oplus\mathbb Z} $$

    The map $\phi:F_2=\langle a,b\rangle\to\mathbb Z\oplus\mathbb Z$ sending $a\mapsto(1,0)$, $b\mapsto(0,1)$ is the abelianisation of $F_2$. Hence its kernel is the commutator subgroup $ [F_2,F_2]= <aba^{-1}b^{-1}> $

    $$ K=\ker\phi=[F_2,F_2]=N, $$ because $N$ is defined as the normal closure of the commutator. No further verification is needed: “kernel of the abelianisation” and “normal subgroup generated by all commutators” are the same object.

     – Sara Apr 23 '25 at 09:32

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Let me first rephrase your question:

Consider the map: $$\begin{aligned} \varphi: F_2&\rightarrow \mathbb{Z}\oplus \mathbb{Z}\\ a &\mapsto (1,0)\\ b &\mapsto (0,1) \end{aligned}$$ Let $K:=\ker\varphi$ and $N:=$ the normal closure of $\{aba^{-1}b^{-1}\}$. You want to prove that $K= N$.

For this purpose, since you have shown $N\subseteq K$, $\varphi$ factors through $$ \begin{aligned} \bar{\varphi}:F_2/N&\rightarrow \mathbb{Z}\oplus \mathbb{Z}\\ aN &\mapsto (1,0)\\ bN &\mapsto (0,1) \end{aligned}$$ and it suffices to prove that $\ker\bar{\varphi}=1\cdot N$ is trivial.

Now take any element in $v\in \ker\bar{\varphi}$. Since it is an element in $F_2/N$ and $abN=baN$, $v$ can be written as the form $a^mb^n N$. Hence: $$ (0,0)=\bar{\varphi}(a^mb^n N)=(m,n)$$ which means $m=n=0$ by construction of $\mathbb{Z}\oplus \mathbb{Z}$. Therefore $v= 1\cdot N$ and $\ker\bar{\varphi}=1\cdot N$ as desired.

Hope this helps.

Zoudelong
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