Question

Let $\mathbf{N}$be a normal subgroup of a group$\mathbf{G}$ and let ${\mathbf{G}}^{\text{'}}$be commutator subgroup defined in Exercise 33. If $\mathit{N}\mathbf{\cap }\mathit{G}\mathbf{\text{'}}\mathbf{=}\mathbf{⟨}\mathbf{e}\mathbf{⟩}$, prove that

.$\mathit{N}\mathbf{\subseteq }\mathit{Z}\mathbf{\left(}\mathbf{G}\mathbf{\right)}$

Short answer

It is proved that,$N\subseteq Z\left(G\right)$ .

Step-by-step solution

Commutator Subgroup

Let $\mathbf{G}$be the group and S be the set of all elements of the form ${\mathbf{aba}}^{-1}{\mathbf{b}}^{-1}$with .$\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{G}$ The subgroup generated by set S is called the commutator subgroup of $\mathbf{G}$.

Let $G$be a group.

Let $N$be a normal subgroup of$G$ and $G\text{'}$be the commutator subgroup of G.

Also, assume that$N\cap G\text{'}=⟨e⟩$

Show that N⊆Z(G)

Let $n\in N$and$g\in G$ be arbitrary elements.

Then it follows from the definition of the commutator subgroup that $ng{n}^{-1}{g}^{-1}\in G\text{'}$.

Also note that since $N$is a normal subgroup of $G$, $ng{n}^{-1}{g}^{-1}\in N$.

Thus,$ng{n}^{-1}{g}^{-1}\in N\cap G\text{'}=\left\{e\right\}$ .

Therefore,$ng{n}^{-1}{g}^{-1}=e$ .

Or equivalently .$ng=gn$

Since both n and g were chosen arbitrarily it follows that,$N\subseteq Z\left(G\right)$ .

Hence proved.