Derived Subgroup is Normal

Theorem
Let $G$ be a group whose identity is $e$.

Let $\sqbrk {G, G}$ denote the derived subgroup of $G$.

Then $\sqbrk {G, G}$ is a normal subgroup of $G$.

Proof
Recall the definition of $\sqbrk {G, G}$:
 * $\sqbrk {G, G}$ is the subgroup of $G$ generated by all its commutators.

Recall also the definition of the commutator of $g, h \in G$:
 * $\sqbrk {g, h} = g^{-1} h^{-1} g h$

From Derived Subgroup is Subgroup we note that $\sqbrk {G, G}$ is indeed a subgroup of $G$.

Let $g, h \in G$ be arbitrary.

Let $x \in \sqbrk {G, G}$ be arbitrary.

By definition of generated subgroup:
 * $\sqbrk {g, h} x \in \sqbrk {G, G}$

and:
 * $\sqbrk {h, g} x \in \sqbrk {G, G}$

Hence also by definition of generated subgroup:
 * $\sqbrk {g, h} x \sqbrk {h, g} \in \sqbrk {G, G}$

and:
 * $\sqbrk {h, g} x \sqbrk {g, h} \in \sqbrk {G, G}$

But from Inverse of Group Commutator:
 * $\sqbrk {g, h}^{-1} = \sqbrk {h, g}$

So:
 * $\sqbrk {g, h} x \sqbrk {g, h}^{-1} \in \sqbrk {G, G}$

and:
 * $\sqbrk {g, h}^{-1} x \sqbrk {g, h} \in \sqbrk {G, G}$

As $x \in \sqbrk {G, G}$ is arbitrary, by definition of coset:


 * $\forall \sqbrk {g, h} \in G: \sqbrk {g, h} \sqbrk {G, G} \sqbrk {g, h}^{-1} \subseteq \sqbrk {G, G}$

and:
 * $\forall \sqbrk {g, h} \in G: \sqbrk {g, h}^{-1} \sqbrk {G, G} \sqbrk {g, h} \subseteq \sqbrk {G, G}$

Hence the result by definition of normal subgroup.