Basic Theorems Regarding the Derived Subgroup of a Group

# Basic Theorems Regarding the Derived Subgroup of a Group

Recall from The Derived Subgroup of a Group page that if $G$ is a group then the derived subgroup of $G$, denoted by $G'$, is the smallest subgroup containing all commutators of $G$.

We will now look at some properties of the derived subgroup of a group.

Proposition 1: Let $G$ be a group. Then:a) $G'$ is a normal subgroup of $G$.b) $G/G'$ is abelian. |

**Proof of a):**Let $g \in G$ and let $h \in G'$. Then $[g, h] \in G'$, i.e., $ghg^{-1}h^{-1} \in G'$. Since $h \in G'$, $ghg^{-1}h^{-1} \in G'$, and $G'$ is a group, we have that:

\begin{align} \quad ghg^{-1} = (ghg^{-1}h^{-1}) \cdot h \in G' \end{align}

- So $gG'g^{-1} \subseteq G'$, and thus, $G'$ is a normal subgroup of $G$. $\blacksquare$

**Proof of b)**Let $g_1, g_2 \in G$ and consider $g_1G', g_2G' \in G/G'$. Since $g_1, g_2 \in G$ we have that $[g_1, g_2] = g_1g_2g_1^{-1}g_2^{-1} \in G$. Thus:

\begin{align} \quad g_1g_2g_1^{-1}g_2^{-1}G' = G' \end{align}

- But also:

\begin{align} \quad g_1g_2g_1^{-1}g_2^{-1}G' = (g_1G')(g_2G')(g_1^{-1}G')(g_2^{-1}G') \end{align}

- Therefore:

\begin{align} \quad (g_1G')(g_2G')(g_1^{-1}G')(g_2^{-1}G' )= G' \end{align}

- Multiplying on the right by $g_2G'$ and then by $g_1G'$ gives us:

\begin{align} \quad (g_1G')(g_2G') = (g_2G')(g_1G') \end{align}

- Since the above equality holds for all $g_1G', g_2G' \in G/G'$ we conclude that $G/G'$ is abelian. $\blacksquare$