Question

Let $\mathit{G}$ be a group of order $\mathit{p}\mathit{q}$, with$\mathit{p}$ and $\mathit{q}$(not necessarily distinct) primes. Prove that the center$\mathit{Z}\mathbf{\left(}\mathbf{G}\mathbf{\right)}$ is either$\mathbf{⟨}\mathbf{e}\mathbf{⟩}$ or$\mathit{G}$ .

Short answer

It is proved that, the center$\mathrm{Z}\left(\mathrm{G}\right)$ of group$G$ is either$\left\{e\right\}$ or$G$ .

Step-by-step solution

Given that

Consider $G$be a group of order $pq$, where $p$and $q$are two primes.

Now, we have to show that $\mathrm{Z}\left(\mathrm{G}\right)$of $\mathrm{G}$is either role="math" localid="1654522190915" $\left\{\mathrm{e}\right\}$or $G$.

It is observed that$\mathrm{Z}\left(\mathrm{G}\right)$is the normal subgroup of$\mathrm{G}$

Assume that $g\in G$and $a\in Z\left(G\right)$, we have $ga{g}^{-1}$as:

$\begin{array}{c}ga{g}^{-1}=g{g}^{-1}a\\ =a\\ \in Z\left(G\right)\end{array}$

This implies that every element in$Z\left(G\right)$ commutes with every element in $G$.

Therefore, the first equality holds.

Prove the result

Assume that $Z\left(G\right)\notin \left\{e\right\}$.

As we know that $Z\left(G\right)$is the normal in $G$, $\mathrm{G}/\mathrm{Z}\left(\mathrm{G}\right)$is a group.

The order of group $G$is $pq$.

Then, by Lagranges theorem, we can say that the order of $Z\left(G\right)$is either $p$or $q$.

Therefore, the order of quotient group $G/Z\left(G\right)$ is either $p$or $q$.

As$p$ and$q$ are prime numbers then, in both cases$G/Z\left(G\right)$ is a cyclic group.

Therefore, G is abelian.

Thus,$\mathrm{Z}\left(\mathrm{G}\right)=\mathrm{G}$.

Hence, it is proved that the center$\mathrm{Z}\left(\mathrm{G}\right)$ of group$G$ is either$\left\{e\right\}$ or $G$.