# The Commutator Subgroup $K$ of $G$ is the “smallest” subgroup such that $G/K$ is Abelian.

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?

• 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. – Bernard Jul 10 '18 at 20:05

$K$ is "smallest" in the sense that any subgroup $H$ with the property that $G/H$ is abelian must contain $K$.
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.