# Derived group

The derived (sub)group (or commutator (sub)group) of a group $G$ is the smallest normal subgroup $D(G)$ of $G$ such that the quotient group $G/D(G)$ is abelian.

Specifically, let $G$ be a group. The group $(G,G)=D(G)$ generated by the set of commutators of $G$ is called the derived group of $G$. It is also called the commutator group of $G$, though in general it is distinct from the set of commutators of $G$. It is a normal subgroup of $G$—in fact, it is a characteristic subgroup.

Evidently, if $H$ is a subgroup of $G$, then $D(H)$ is a subgroup of $D(G)$.

Proposition. Let $G$ and $G'$ be groups, and $f: G \to G'$ be a group homomorphism. Then $f(D(G)) \subseteq D(G')$; in fact, $f(D(G)) = D(f(G))$, so in particular, if $f$ is surjective, $f(D(G)) = D(G')$.

Proof. If $(a,b)$ is a commutator of $G$, then $f((a,b)) = (f(a),f(b))$ is a commutator of $G'$; thus $f(D(G)) \subseteq D(G')$. Suppose $x,y$ are points in the image of $G$ under $f$; let $a,b$ be elements of $G$ such that $f(a)=x, f(b)=y$. Then $f(a,b) = (f(a),f(b)) = (x,y)$. Hence $f(D(G)) = D(f(G))$. $\blacksquare$

Corollary 1. The derived group $D(G)$ is a characteristic subgroup of $G$. In particular, it is normal.

Corollary 2. The quotient group $G/D(G)$ is commutative. Let $\lambda$ be the canonical homomorphism from $G$ to $G/D(G)$. Let $G'$ be an abelian group. Then every homomorphism $f: G \to G$ can be expressed uniquely as $f' \circ \pi$, where $f' : G/D(G) \to G'$ is a homomorphism.

Proof. Note that $D(G/D(G)) = D(\lambda(G)) = \lambda(D(G)) = \{e\}$; thus $G/D(G)$ is abelian. If $f:G\to H$ is a homomorphism, then $f(D(G)) \subseteq D(H) = \{e\}$. Thus the realtion $f(a) = f(b)$ is compatible with equivalence mod $D(G)$. $\blacksquare$

Corollary 3. Let $H$ be a subgroup of $G$. Then $D(G) \subseteq H$ if and only if $H$ is normal in $G$ and $G/H$ is commutative.

Proof. If $H$ contains $D(G)$, then $H/D(G)$ is normal in $G/D(G)$, since every subgroup of an abelian group is normal; hence $H$ is normal, and $G/H$ is isomorphic to $(G/D(G))/(H/D(G))$, which is commutative. The converse follows from the previous corollary.

Corollary 4. Let $X$ be a generating subset of $G$. Then $D(G)$ is the normal subgroup generated by the set of commutators of elements of $X$.

Proof. Let $H$ be the normal subgroup generated by the commutators of elements of $X$. By definition, $H \subseteq D(G)$. On the other hand, the set $X/H$ generates the group $G/H$; since the elements of $X/H$ commute, $G/H$ is abelian, and hence $D(G) \subseteq H$. $\blacksquare$