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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.

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    $\begingroup$ You already have it all: if the quotient $\,G/Z(G)\,$ cannot be cyclic non-trivial then... $\endgroup$ – DonAntonio May 12 '13 at 3:44
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    $\begingroup$ @DonAntonio Perhaps undelete your answer so this doesn't get listed as unanswered. $\endgroup$ – Ragib Zaman May 12 '13 at 6:42
  • $\begingroup$ Done @Ragib, thanks. $\endgroup$ – DonAntonio May 12 '13 at 10:06
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Hint:

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

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