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If $\vert G\vert=pq$ with $p$ and $q$ prime numbers, then either $G$ is abelian or $Z(G) = \{ e\}$. What are the options for the order of $G/Z(G)$?

So I know that if $\vert Z(G)\vert = 1$, then $Z(G) = {e}$. And to prove that $G$ is abelian when $Z(G)$ is not just the identity element, I know that if $\vert G/Z(G)\vert \in \{p,q\}$, it would imply that $G/Z(G)$ is cyclic because $p$ and $q$ are prime numbers, and $G$ would be abelian (I've already proven that bit); but I am not quite sure on how to determine the possible orders of the quotient group, so that I could use the previous argument.

And we have not yet learned about centralizer and class equation and those things so I can't use them.

Any help or hints are very much appreciated, thank you.

Shaun
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Tomas Rojas
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  • You're almost there. If $G$ is abelian, what do you know abut $Z(G)$? – Robert Shore Apr 07 '23 at 00:44
  • well if $G$ is abelian then $G=Z(G)$ and that implies that $G = G/Z(G)$ right? and $\vert G/Z(G)\vert = \vert G\vert = pq$ too. But I am still confused, is $pq$ the only option for the order of $G/Z(G)$? and if It is then I know that $G$ must be abelian? Sorry for so many questions, I am still a bit confused with this topic. – Tomas Rojas Apr 07 '23 at 01:56
  • Not quite. If $G=Z(G)$, then $G/Z(G)=G/G$, which has order $1$. – Robert Shore Apr 07 '23 at 02:00
  • If the center is the whole group, then there is one coset only of the center in the group, namely the center itself. So, $Z(G)=G\Longrightarrow |G/Z(G)|=1$. What happens if $|Z(G)|=p$ or $q$? – Kan't Apr 07 '23 at 02:26
  • (cont'd) How many cosets you'd have accordingly? Those are the sizes of $|G/Z(G)|$. – Kan't Apr 07 '23 at 02:33

2 Answers2

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By Lagrange's Theorem, since $Z(G)$ is a subgroup of $G$ (always), we have that $|Z(G)|\in \{1, p, q, pq\}$.

If $|Z(G)|=p$, then $|G/Z(G)|=\frac{pq}{p}=q$, a prime. Since all groups of prime order are cyclic, this means that $G/Z(G)$ is cyclic. But then, $G$ is abelian (have you proven this?).

Similarly, if $|Z(G)|=q$, then $|G/Z(G)|=p$, so again $G$ is abelian.

Why does it matter that $G$ is abelian? Well, if $G$ is abelian, then all of its elements commute with everything, so $G=Z(G)$. That means that $|Z(G)|=|G|=pq$, a contradiction. Thus $|Z(G)|\neq p, q$.

This leaves us with two options: $|Z(G)|=1$, $|Z(G)|=pq$.

If $|Z(G)|=1$ then $Z(G)=\{e\}$ (all subgroups contain the identity).

If $|Z(G)|=pq$, then $|Z(G)|=pq=|G|$ so $Z(G)=G$ (they have the same elements).

Thus either $G/Z(G)=G/\{e\}\cong G$ or $G/Z(G)=G/G\cong \{e\}$.

Hence, $|G/Z(G)|\in \{1, pq\}$.

IAAW
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  • To clarify in case there is any confusion, if $G$ is a finite group and $H$ is a subgroup of $G$, by Lagrange's Theorem, $|G/H|=\frac{|G|}{|H|}$. In fact, this is the main result of Lagrange's Theorem. – IAAW Apr 09 '23 at 00:18
  • hey thank you for your answer. really helpful. I only got one question about the contradiction, why is |()|=||= a contradiction? I don't get why |()| can't be pq. – Tomas Rojas Apr 13 '23 at 23:37
  • No problem! It's a contradiction because we assumed $|Z(G)|=p, q$. Of course, $|Z(G)|$ cannot be equal to both $p$ and $pq$. – IAAW Apr 13 '23 at 23:41
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Hint: If $G$ is a finite group, then for any subgroup $H$ of $G$, the order of $H$ divides the order of $G$.

C Squared
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