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A nonabelian group $G$ of order $pq$ where $p$ and $q$ are primes has a trivial center

My Proof is as follows:

Assume we have nonabelian group $G$ of order $pq$ where both $p$ and $q$ are primes. When $G$ has a trivial center it means subgroup $Z(G)=\{e\}$. If a group is of order $pq$ then the order of its subgroup must divide pq meaning $Z(G)$ has to be of order $1$, $p$ or $pq$. It cannot be $pq$ otherwise $Z(G)$ would be a group equivalent to $G$ and not a subgroup. It cannot be $p$ or $q$ because otherwise $Z(G)$ would be cyclic and therefore abelian. So $Z(G)$ must be or order $1$.

Is this correct? A hint I was given was to use the fact that if $G/Z(G)$ is abelian then it is cyclic. How would that be incorporated?

cambelot
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    Are you sure that the hint is not actually: If $G/Z(G)$ is cyclic then $G$ is abelian"? – N. S. Oct 31 '14 at 01:45
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    If $Z(G)$ has order $pq$, $Z(G)$ is still a subgroup. For any group $G$, $G$ is a subgroup of $G$. There's a different reason $Z(G)$ can't have order $pq$. – Stahl Oct 31 '14 at 01:46
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    The proof is full of errors ... – Martin Brandenburg Oct 31 '14 at 09:12
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    It seems the answers are all focused on giving a correct proof rather than pointing out the main error in your approach, so allow me to do it. The fact that $Z(G)$ would be abelian when its order were $p$ or $q$ is not really a useful argument since we know that $Z(G)$ is abelian from the start, no matter what its order turns out to be. It is in the nature of being the center that $Z(G)$ is abelian. – Vincent Apr 10 '21 at 21:10
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    +1 for @Vincent, someone should have pointed out the obvious flaw. – Paramanand Singh Apr 11 '21 at 03:25

4 Answers4

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I think the correct hint is:

If $G/Z(G)$ is cyclic then $G$ is abelian.

$Z(G) \subseteq G$ and so we can have $$|Z(G)| = 1, \ p , \ q , \ pq $$ $G$ is nonabelian and so $|Z(G)| \neq pq $.

If $|Z(G)| = p $ or $|Z(G)| = q $ the quotient group $G/Z(G)$ has prime order, whence is cyclic and by the hint $G$ is abelian.

Thus $|Z(G)| = 1$

WLOG
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Your result can be proved directly:

Suppose $Z(G) \neq \{1 \}$. Because $G$ is not abelian, $Z(G)$ has order $p$ or $q$; say $|Z(G)|=p$. In particular, there exists $x \in Z(G)$ of order $p$. Let also $y \in G$ of order $q$. Now, it is easy to notice that the set $$X= \{ x^m y^n \mid 1 \leq m \leq p, \ 1 \leq n \leq q \}$$ has cardinality $pq$, hence $G=X$. Now, because $x \in Z(G)$, we clearly deduce that $G$ is abelian: a contradiction.

Seirios
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Let's assume $|Z(G)|=q$; then $|C_G(g)|=q$ for every noncentral $g\in G$, whence $\frac{|G|}{|C_G(g)|}=\frac{pq}{q}=p$ for every noncentral $g\in G$. From the Class Equation ($pq=q+q(p-1)$), follows: $q(p-1)=kp$, where $k$ is the number of noncentral conjugacy classes of $G$: contradiction, because $p\nmid q(p-1)$. The same argument holds by swapping $p$ and $q$, and hence $|Z(G)|\ne p$ as well. As $G$ is nonabelian, we are left with $Z(G)=1$.

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More generally, let $G$ be of order $pq$. If it has center of 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 $k=q$, because $C_G(g)\le G$ and 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 $pq$. So, if $G$ is assumed nonabelian, it must be centerless.