The Derived Subgroup of a Group

# The Derived Subgroup of a Group

Recall from The Commutator of Two Elements in a Group page that if $G$ is a group and $g_1, g_2 \in G$ then the commutator of $g_1$ and $g_2$ is defined to be:

(1)
\begin{align} \quad [g_1, g_2] = g_1g_2g_1^{-1}g_2^{-1} \end{align}

In general, the set of commutators in $G$, $\{ [g_1, g_2] : g_1, g_2 \in G \}$ might not be a subgroup of $G$. That said, there always exists a smallest group containing the set of all commutators (namely, the intersection of all subgroups containing the set of commutators), and we give this subgroup a name.

 Definition: Let $G$ be a group. The corresponding Derived Subgroup of $G$ denoted by $G'$ or $[G, G]$ is the smallest subgroup of $G$ containing all commutators of $G$.

We now look at some properties of the derived subgroup $G'$ of a group $G$.

 Proposition 1: Let $G$ be a group. Then $G$ is abelian if and only if $G'$ is the trivial group.
• Proof: $\Rightarrow$ Suppose that $G$ is abelian. Then for all $g_1, g_2 \in G$ we have that $g_1g_2 = g_2g_1$. So $g_1g_2g_1^{-1}g_2^{-1} = e$ for all $g_1, g_2 \in G$ where $e$ is the identity element of $G$.
• In otherwords, $[g_1, g_2] = e$ for all $g_1, g_2 \in G$. Clearly the smallest subgroup of $G$ containing all the commutators is the trivial group. Thus $G'$ is the trivial group.
• $\Rightarrow$ Suppose that $G'$ is the trivial group. Then for all $g_1, g_2 \in G$ we have that $[g_1, g_2] = e$. That is, for all $g_1, g_2 \in G$ we have that $g_1g_2g_1^{-1}g_2^{-1} = e$. So $g_1g_2 = g_2g_1$ for all $g_1, g_2 \in G$, i.e., $G$ is abelian. $\blacksquare$
 Definition: Let $G$ be a group. The Second Derived Subgroup of $G$ is defined to be $G^{(2)} = (G')'$, that is, $G^{(2)}$ is the smallest subgroup of $G'$ containing all commutators of $G'$. If $n \in \mathbb{N}$ we defined the $n^{\mathrm{th}}$ Derived Subgroup of $G$ to be defined as $G^{(n)} = (G^{(n-1)})'$, and it is the smallest subgroup of $G^{(n-1)}$ containing all commutators of $G^{(n-1)}$.

From the definition above, we see that if $G$ is a group and $n \in \mathbb{N}$ then:

(2)
\begin{align} \quad G^{(n)} \leq G^{(n-1)} \leq ... \leq G^{(1)} \leq G \end{align}