Let $G$ be a group of order $pq$, with $p$ and $q$ prime. Prove that the order of the center of $G$ is 1 or $pq$.

Let me start off with what I did:

Assume $G$ is abelian. Then we know $Z(G)=G$, thus the center must have order $pq$.

Now assume $G$ is non-abelian. By Cauchy's theorem we know there must exist elements $x$ and $y$ of respectively order $p$ and $q$. Without loss of generality we can assume $p \leq q$. Because $[G: \langle y \rangle ] = p$, and $p$ is the smallest prime divisor of $G$, $\langle y \rangle$ must be a normal subgroup of $G$.

I claim that $\langle x \rangle$ is a complete set of representatives for $\langle y \rangle$ in $G$. We know, because $p$ is prime, that $\langle x \rangle$ is cyclic and therefore, $\langle x \rangle = \{ x^k : k \in \{ 0, \dots , p-1 \} \}$. We can assume $i < j$. If $x^i \langle y \rangle = x^j \langle y \rangle$, then $x^{-i}x^j = x^{j-i} \in \langle y \rangle$. But $x^{j-i} \in \langle x \rangle$, thus of order p, hence $x^{j-i}$ can't be in $\langle y \rangle$. As so, $\langle x \rangle$ gives a full set of representatives.

Now we can write every element in $G$ as $x^m y^n$, with $m \in \{ 0 , \dots , p-1\}$ and $n \in \{ 0, \dots , q-1\}$. Furthermore, every element other than the identity element has either order $p$ or $q$, or else $G$ would be cyclic, and thus abelian.

Now I'm kind of stuck, but I want to prove that there exists an element in $G$ which doesn't commute with anything in $G$, except for the identity element.

I know there is an easier solution to this, namely by saying that the order of $Z(G)$ must divide the order of the group $G$. So either it must be $1, p, q$ or $pq$. However it can't be $pq$, or else the group would be cyclic and thus abelian. Nor can it be $p$, since then $[G : Z(G) ] = q$, and thus it's cyclic, so by a specific theorem $G$ must be abelian. The same argument for $q$.

I think the proof of that specific theorem has to do with the proof I started. However I can't connect the dots. But I do want to know how I should finish it this way (if that is possible).

Thanks in advance for any help.

  • 9
    $\begingroup$ You already have it all: if the quotient $\,G/Z(G)\,$ cannot be cyclic non-trivial then... $\endgroup$
    – DonAntonio
    May 12, 2013 at 3:44
  • 1
    $\begingroup$ @DonAntonio Perhaps undelete your answer so this doesn't get listed as unanswered. $\endgroup$ May 12, 2013 at 6:42
  • $\begingroup$ Done @Ragib, thanks. $\endgroup$
    – DonAntonio
    May 12, 2013 at 10:06

2 Answers 2



For any group $\,G\,$ , the quotient group $\,G/Z(G)\,$ cannot be cyclic non-trivial.


If the center has order $p$, then every noncentral element $g$ has centralizer of order $kp$ for some integer $k>1$, because $Z(G)<C_G(g)$, and then necessarily $k=q$, for the primality of $q$. But then $C_G(g)=G$, and hence $g\in Z(G)$: contradiction. Same argument and conclusion if the center has order $q$. Therefore, the center has order $1$ or $pq$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.