Question

Let $\mathit{G}$ be a group and let $\mathit{S}$ be the set of all elements of the form $\mathit{a}\mathit{b}{\mathit{a}}^{\mathbf{-}\mathbf{1}}{\mathit{b}}^{\mathbf{-}\mathbf{1}}$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 proven that $G^{\text{'}}$is normal in $G$.

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 a subgroup of $G$ generated by the set $S$.

Prove the result

Consider the elements as $a,b\in G$ and $g\in G$ such that $g\left(ab{a}^{-1}{b}^{-1}\right)g^{-1}\in G^{\text{'}}$.

Simplify as:

$\begin{array}{c}g\left(ab{a}^{-1}{b}^{-1}\right)g^{-1}=ga{g}^{-1}gb{g}^{-1}g{a}^{-1}{g}^{-1}g{b}^{-1}{g}^{-1}\\ =\left(ga{g}^{-1}\right)\left(gb{g}^{-1}\right)\left(g{a}^{-1}{g}^{-1}\right)\left(g{b}^{-1}{g}^{-1}\right)\end{array}$

This implies that for any $s\in S$, and $g\in G$, then, we have $gs{g}^{-1}\in G^{\text{'}}$.

Apply the definition of $G^{\text{'}}$, consider any elements as $x\in G^{\text{'}}$ and $g\in G$, then there exists $s_1,\dots ,s_k\in S$ and $i_1,\dots ,i_k\in \mathbb{N}$, such that $x=s_1^{i_1}\dots s_k^{i_k}$.

This implies that:

$$\begin{gathered} gx{g}^{-1}=g{s}_1^{i_1}\dots s_k^{i_k}{g}^{-1} \\ =g{s}_1^{i_1}{g}^{-1}g\dots g^{-1}g{s}_k^{i_k}{g}^{-1} \\ ={\left(g{s}_1{g}^{-1}\right)}^{i_1}{\left(g{s}_2{g}^{-1}\right)}^{i_2}\dots {\left(g{s}_{k-1}{g}^{-1}\right)}^{k-1}{\left(g{s}_k{g}^{-1}\right)}^{i_k} \\ \in G^{\text{'}} \end{gathered}$$

Therefore, it is observed that $g{G}^{\text{'}}{g}^{-1}\subset G^{\text{'}}$ for any $g\in G$.

Hence, by using the theorem, $G^{\text{'}}$ is a normal subgroup.