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 $G$.

Recall the definition of $\sqbrk {G, G}$:

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.

$\sqbrk {g, h} x \in \sqbrk {G, G}$

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:

$\sqbrk {h, g} x \sqbrk {g, h} \in \sqbrk {G, G}$

$\sqbrk {g, h}^{-1} = \sqbrk {h, g}$

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.

$\blacksquare$