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Let $p$ and $q$ primes, with $p<q$ and $pk \neq q-1 \ , \ \forall k \in \mathbb{Z}$. Show that there does not exist group $G$ such that $$\left|\frac{G}{Z(G)}\right|=pq,$$ where $Z(G)$ is the center of $G$.

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Hint: prove the basic

Lemma: For any group $\,G\,$ , the quotient $\,G/Z(G)\,$ cannot be cyclic non-trivial, or in other words: for any group, $\,G/Z(G)\,$ is cyclic iff $\,G\,$ is abelian, and in this case the quotient is the trivial group.

Now just show that under the given data, a group of order $\,pq\,$ must be cyclic...

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continue please –  Jarbas Dantas Silva Oct 17 '12 at 1:16
    
What "continue" and where? –  DonAntonio Oct 17 '12 at 2:55
    
Why a group of order pq must be ciclic? –  Jarbas Dantas Silva Oct 17 '12 at 15:57
    
(1) How many Sylow $\,p-\,$subgroups it has? (2) Thus, what is $\,PQ\,$ , with $\,P,Q\,$ Sylow subgroups of order $\,p,q\,$ respectively? (3) Thus, $\,PQ\cong P\times Q\,$ ... –  DonAntonio Oct 17 '12 at 17:05
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