Question

Let $\mathit{G}$ be a group and let $\mathit{S}$ be the set of all elements of the form role="math" localid="1654524634371" $\mathit{a}\mathit{b}{\mathit{a}}^{\mathbf{-}\mathbf{1}}{\mathit{b}}^{\mathbf{-}\mathbf{1}}$with $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{G}$. The subgroup$\mathit{G}\mathbf{\text{'}}$ generated by the set $\mathit{S}$(as in Theorem 7.18) is called the commutator subgroup of $\mathit{G}$. Prove

1. ${\mathit{G}}^{\mathbf{\text{'}}}$is normal in$\mathit{G}$ .
2. $\mathit{G}\mathbf{/}{\mathit{G}}^{\mathbf{\text{'}}}$is abelian.

Short answer

It is proved that $G/G^{\text{'}}$ is an abelian group.

Step-by-step solution

Define the sets

Consider $G$ to be a group and $S$be the subset as $S=\{ab{a}^{-1}{b}^{-1}:a,b\in G\}$.

Also, consider $G\text{'}$ to be the subgroup of $G$ generated by the set $S$.

Prove the result

We have $G^{\text{'}}$ as a normal subgroup; this implies that $G/G^{\text{'}}$ is a group. Assume that$g{G}^{\text{'}},h{G}^{\text{'}}\in G/G^{\text{'}}$ be any two arbitrary elements. Then we have $gh{g}^{-1}{h}^{-1}\in G^{\text{'}}$for any $g,h\in G$.

Therefore,$gh{G}^{\text{'}}=hg{G}^{\text{'}}$ .

This implies that $G/G^{\text{'}}$ is an abelian group.