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Let $G$ be the free group on two generators, and $K$ its commutator. We have that $G/K \cong \mathbb{Z} \oplus \mathbb{Z}$. This isomorphism is induced by $\theta: G \to \mathbb{Z} \oplus \mathbb{Z}$ given by $s \mapsto (\#a(s), \#b(s))$, where $\#a(s)$, $\#b(s)$ is the number of symbols "$a$" respectively "$b$" in the string $s$ (it can be also negative). For instance, $\theta(a^{-1}ba^2b^{-3}a^2) = (3, -2)$. It is evident that all elements of the commutator are mapped to $(0, 0)$. Therefore, $\text{Ker}\,\theta \supseteq K$. But how do I see the opposite direction, i.e. $\text{Ker}\,\theta \subseteq K$?

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  • $\begingroup$ You mean "$K$ is its commutator subgroup" not just "$K$ is its commutator". $G/K$ is abelian, so the images in $G/K$ of any element in which the exponent sums of both $a$ and $b$ are zero is trivial. Hence $\ker \theta \le K$. $\endgroup$ – Derek Holt Oct 11 '15 at 17:27
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I think what you mean is that $K$ is the group generated by commutators. For any $s∈Ker(θ)$,we know the sum of the number(including nagative) of "a" and "b" are zero. Without loss of generality,we can write $$s=a^{n_1}b^{m_1}a^{n_2}b^{m_2}…a^{n_t}b^{m_t}$$,with $$\sum_{i=1}^tn_i=\sum_{i=1}^tm_i=0$$ The other three forms can be proved similarly. Now we can see that $$s=a^{n_1}b^{m_1}a^{n_2}b^{m_2}…a^{n_t}b^{m_t}=a^{n_1}b^{m_1}a^{-n_1}b^{-m_1}b^{m_1}a^{n_1+n_2}b^{m_2}…a^{n_t}b^{m_t}$$$$=c_1·b^{m_1}a^{n'_2}b^{m_2}…a^{n_t}b^{m_t}$$ $$c_1=a^{n_1}b^{m_1}a^{-n_1}b^{-m_1}∈K $$$$n'_2=n_1+n_2$$ Terms after $c_1$still satisfies $$\sum_{i=3}^tn_i+n'_2=\sum_{i=1}^tm_i=0$$ We can see that the number of terms except $c_i$'s minus at lest one if we do this process.So repeat it until $s$ becomes a multiplication of commutators $c_i$ ,since t is finite for each s.Thus $Ker(θ)⊆K$.

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