# If $|G|=pq$ where $p,q$ are primes that are not necessarily distinct, prove the order of $Z(G)=1$ or $pq$.

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.

• Hint: prove that if a group $G$ satisfies that $G/Z(G)$ is cyclic, then $G$ is abelian. Jul 14, 2016 at 14:00

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 $1$.

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 $$1$$.

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.

• 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 $n\geq 3$. Jul 14, 2016 at 16:17