6
$\begingroup$

Question : If $|G|=pq$ where $p$ and $q$ are primes that are not necessarily distinct. Prove that the order of $Z (G) =1$ or $pq$.

Showing the order is $pq$ is trivial.

I unsure how to start with showing the order is $1$. Hints are appreciated.

Thanks in advance.

$\endgroup$
1
  • 2
    $\begingroup$ Hint: prove that if a group $G$ satisfies that $G/Z(G)$ is cyclic, then $G$ is abelian. $\endgroup$
    – PseudoNeo
    Jul 14, 2016 at 14:00

3 Answers 3

8
$\begingroup$

By Lagrange theorem, $|Z(G)|=1,p,q$ or $pq$.

If $|Z(G)|=p$ or $|Z(G)|=q$ the quotient group $G/Z(G)$ has prime order (respectively $q$ or $p$). So $G/Z(G)$ is cyclic, and so $G$ is abelian (proof here). But if $G$ is abelian, then $Z(G)=G$, so $|Z(G)|=pq$. So there is a contradiction.

Finally : $|Z(G)| = pq$ or $1$.

$\endgroup$
2
$\begingroup$

Use the fact that $Z(G)$, i.e. the center of $G$, is a subgroup of $G$, so by Lagrange Theorem $\big| Z(G) \big| \bigg| \big| G \big|$, hence $\big| Z(G) \big| = \{1,p,q,pq\}$. If $\big| Z(G) \big| = p$ or $q$, then the order of $G/Z(G)$ is a prime number, hence the quotent group is cyclic, implying that $G$ is abelian. A contradiction. Therefore $\big| Z(G) \big| = \big| G \big| = pq $ or $1$.

$\endgroup$
1
$\begingroup$

Suppose not, and let $x$ be an element in the center, without loss of generality assume it is of order $p$, by cauchy's theorem there is an element $y$ of order $q$. So $x$ and $y$ commute and generate $G$. $G$ is therefore abelian.

Note that this can be generalized as follows: Let $G$ be a non-abelian group such that for every $x\in G\setminus\{e\}$ there is $y$ with $\langle x,y\rangle=G$, then the center of $G$ is trivial.

$\endgroup$
1
  • $\begingroup$ A corollary of this is that $S_n$ has trivial center for $n\neq 4$. Of course, showing that $S_n$ has the property that for every $x\neq e$ there is a $y$ so that $\langle x,y \rangle =S_n$ is much harder than proving the center of $S_n$ is trivial for $n\geq 3$. $\endgroup$
    – Asinomás
    Jul 14, 2016 at 16:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .