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Let $G$ be a group. For any $a,b\in G$ , the element $a^{-1}b^{-1}ab$ is called the commutator of $a$ and $b$ .
The commutator $a^{-1}b^{-1}ab$ is sometimes written $[a,b]$ . (Usage varies, however, and some authors instead use $[a,b]$ to represent the commutator $aba^{-1}b^{-1}$ .) If $A$ and $B$ are subsets of $G$ , then $[A,B]$ denotes the subgroup of $G$ generated by $\{[a,b]\mid a\in A{ and }b\in B\}$ . This notation can be further extended by recursively defining $[X_1,\dots,X_{n+1}]=[[X_1,\dots,X_n],X_{n+1}]$ for subsets $X_1,\dots,X_{n+1}$ of $G$ .
The subgroup of $G$ generated by all the commutators in $G$ (that is, the smallest subgroup of $G$ containing all the commutators) is called the derived subgroup, or the commutator subgroup, of $G$ . Using the notation of the previous paragraph, the derived subgroup is denoted by $[G,G]$ . Alternatively, it is often denoted by $G'$ , or sometimes $G^{(1)}$ .
Note that $a$ and $b$ commute if and only if the commutator of $a,b\in G$ is trivial, i.e.,$$ a^{-1} b^{-1}a b = 1.$$ Thus, in a fashion, the derived subgroup measures the degree to which a group fails to be abelian.
The factor group $G/[G,G]$ is the largest abelian quotient of $G$ , and is called the abelianization of $G$ .
One can of course form the derived subgroup of the derived subgroup; this is called the second derived subgroup, and denoted by $G''$ or $G^{(2)}$ . Proceeding inductively one defines the $n\supth$ derived subgroup $G^{(n)}$ as the derived subgroup of $G^{(n-1)}$ . In this fashion one obtains a sequence of subgroups, called the derived series of $G$ : $$G=G^{(0)} \supseteq G^{(1)} \supseteq G^{(2)} \supseteq \cdots$$
Proposition 2 The group $G$ is solvable if and only if the derived series terminates in the trivial group $\{ 1 \}$ after a finite number of steps.
The derived series can also be continued transfinitely--see the article on the transfinite derived series.
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