Derived Subgroup is Subgroup

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 indeed a 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 $g, h \in G$:

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

We note that from Commutator of Group Element with Identity is Identity:

$\sqbrk {e, e} = e$

and so $e \in \sqbrk {G, G}$

Thus:

$\sqbrk {G, G} \ne \O$

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

We have by definition of inverse element that:

$\sqbrk {g, h} = \sqbrk {g, h} e$

and so $\sqbrk {g, h} \in \sqbrk {G, G}$.

Let $g_1, h_1, g_2, h_2 \in G$ be arbitrary.

By definition of $\sqbrk {G, G}$:

$\sqbrk {g_1, h_1} \in \sqbrk {G, G}$

and also:

$\sqbrk {g_2, h_2} \in \sqbrk {G, G}$

Hence by definition of generated subgroup:

$\sqbrk {g_1, h_1} \sqbrk {g_2, h_2} \in \sqbrk {G, G}$

Hence by the Two-Step Subgroup Test, $\sqbrk {G, G}$ is a subgroup of $G$.

$\blacksquare$