Question

Let $\mathit{N}$ be a normal subgroup of a group$\mathit{G}$ 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

The center of$\mathit{G}\mathbf{/}\mathit{N}$ is$\mathit{Z}\mathbf{\left(}\mathbf{G}\mathbf{\right)}\mathbf{/}\mathit{N}$ .

Short answer

It is proved that, the center of$G/N$ is $Z\left(G\right)/N$.

Step-by-step solution

Commutator Subgroup

Let $\mathit{G}$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 localid="1654526981526">G'generated by set S is called the commutator subgroup of $\mathbf{G}$.

Let $G$be a group.

Let $N$ be a normal subgroup of localid="1654527130633" $G$and localid="1654527276660" $G\text{'}$be the commutator subgroup of G.

Also, assume that localid="1654527292959" $N\cap G\text{'}=⟨e⟩$.

Show that center of G/N is Z(G)/N

Since $N$is a normal subgroup of $G$and $N\subseteq Z\left(G\right)$, $N$is also a normal subgroup of $Z\left(G\right)$.

Now, note that since every element in $Z\left(G\right)$commutes with every element of $G$, it follows that every element in $Z\left(G\right)/N$commutes with every element of $G/N$.

Thus,$Z\left(G\right)/N$ is contained in the centre of $G/N$.

On the other hand, let $\left[g\right]$ be an element in the centre of $G/N$.

Then, we show that $g\in Z\left(G\right)$.

Since $\left[g\right]$is an element in the centre of $G/N$, $\left[g\right]\left[h\right]=\left[h\right]\left[g\right]$for all $\left[h\right]\in G/N$.

Or equivalently, $\left[gh{g}^{-1}{h}^{-1}\right]=e\text{in}G/N$.

Consequently,$gh{g}^{-1}{h}^{-1}\in N$but$gh{g}^{-1}{h}^{-1}\in G\text{'}$.

As well, it follows from the hypothesis that,

$gh{g}^{-1}{h}^{-1}=e$

Thus,$gh=hg$ in G.

For all$h\in G$implying that$g\in Z\left(G\right)$, which verifies the claim.

Therefore, we have shown that the centre of$G/N$is contained in $Z\left(G\right)/N$.

Thus, we have that the centre of $G/N$ is $Z\left(G\right)/N$.