If $G$ be a group of order $pq$ where $p$ and $q$ are prime integers then find $|Z(G)|$. The options are i) 1 or $p$ ii) 1 or $q$ iii) 1 or $pq$ iv) None of these. I know groups of prime order are cyclic so $G$ will atleast have one subgroup of order $p$ or $q$. But how will I find out $|Z(G)|$ from this?
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Are $p$ and $q$ distinct? – Stefan Apr 02 '15 at 11:43
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@Stefan - That does not matter. – Nicky Hekster Apr 02 '15 at 12:00
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1@NickyHekster: If $p=q$ the group is necessarily abelian and so $Z=G$. If $p\ne q$ then the group may be non abelian(dihedral). – Apr 02 '15 at 12:08
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@Shabab - yes exaclty so answer (iii) applies. – Nicky Hekster Apr 02 '15 at 12:18
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You will need the result that if $G/Z(G)$ is cyclic, then $G$ is abelian. This rules out $\lvert Z(G) \rvert=p$ or $q$. If $G$ is abelian, then $G=Z(G)$. Also, $\lvert Z(G) \rvert=1$ can be realized (an example is $G=S_3$, but can be generalized to $p$ or $q$ arbitrary, with $p \neq q$).
Nicky Hekster
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How do I know if $G/Z(G)$ is cyclic or not? I could not understand why are you ruling out $|Z(G)| = p$ or $q$. – In78 Apr 02 '15 at 16:41
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@In78 - $|Z(G)| \mid |G|$. So $|Z(G)| \in { 1, p, q, pq }$. If $|Z(G)|=p$, then $|G/Z(G)|=q$, so $G/Z(G)$ is cyclic, whence $G$ is abelian and $G=Z(G)$, a contradiction. Same reasoning applies to the case $|Z(G)|=q$. Hope this clarifies it. – Nicky Hekster Apr 03 '15 at 08:09