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Let $G$ be a group. A commutator is an element of the form $aba^{^-1}b^{-1}$. The set of finite products of commutators is a normal subgroup $K$ called the commutator subgroup.

The book claims $K$ is the smallest subgroup such that the quotient $G/K$ is abelian.

I'm wondering what they mean by smallest. Is it "smallest" in order? If so then the quotient $G/K$ should have the "largest" order out of the possible quotient groups of $G$ that are abelian. Is this what is meant?

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    $\begingroup$ This means the smallest for inclusion. In other words, a normal subgroup $H$ is such that $G/H$ is abelian if and only if it contains the commutators. $\endgroup$ – Bernard Jul 10 '18 at 20:05
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$K$ is "smallest" in the sense that any subgroup $H$ with the property that $G/H$ is abelian must contain $K$.

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Suppose that $G/K$ is Abelian. Note that

$$\forall a,b \in G: abK = baK \iff \forall a,b\in G:a^{-1}b^{-1}ab \in K$$ This means that $\{[a,b]: a,b \in G\} \subseteq K \subseteq G$ and hence $G'=\langle \{[a,b]: a,b \in G\} \rangle \subseteq K$.

That means that $G'$ is the smallest subgroup of $G$ such that the quotient group is Abelian.

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