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I came across this problem that says:

If $Z(G)$ denotes the centre of the group $G$, then the order of the quotient group $G/Z(G)$ can not be which of the following?

(a) $15,$
(b) $25,$
(c) $6,$
(d) $4.$

Could someone point me in the right direction (a certain theorem or property that I have to use)? Thanks everyone in advance for your time.

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5  
It's a well-known fact (which you should prove if you haven't seen it) that $G/Z(G)$ is never a nontrivial cyclic group. –  Chris Eagle Dec 6 '12 at 14:47
    
If the quotient is cylcic, then $G$ is itself abelian. Moreover, if m is prime to n, then there is only one extension of $Z_m$ by $Z_n$. The later fact could be seen, either by constructing an isomorphism between any two extensions, or by the fact that Ext is the derived functor of Hom, together with the fact that, if $G$ is a $p$-group and $A$ a $q$ torsion module, then the first cohomology group vanishes. This serves only as a reference, for we can derive the same result by simple group-theoretical considerations, of course. –  awllower Dec 6 '12 at 15:10
    
My comment above attempted to show that $Z_{15}$ is cyclic. But we can always show this by counting its numbers of Sylow-subgroups, which are all=1, hence it is cyclic. –  awllower Dec 6 '12 at 15:17
    
But your comment about when $m$ and $n$ are coprime is not correct. For example, there are two groups of order $6$. –  Tobias Kildetoft Dec 6 '12 at 15:41
    
@Tobias Indeed it was wrong, but I cannot figure out why... Could you please tell me where my arguments failed? Thanks in any case. –  awllower Dec 6 '12 at 16:23
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1 Answer 1

up vote 3 down vote accepted

We have the following theorem (a consequence of Sylow's theorems):

Let $G$ be a group of order $pq$ where $p,q$ are primes such that $p < q$ and $p$ does not divide $q-1$. Then $G$ is cyclic.

If $G / Z(G)$ has order $15$ then $p=3$ and $q-1 = 5-1 = 4$ so that $p$ does not divide $q-1$. Hence choice (a) is not possible using Chris Eagle's comment:

If $G / Z(G)$ is cyclic it follows that $G$ is abelian (prove it) so that $Z(G) = G$ and hence $G/Z(G) = \{0\}$.

Hope this helps.

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If G/Z(G) has order 6 then p=2 and q−1=3−1=2 so that p does divide q−1. Hence choice (c) is possible using Chris Eagle's comment. –  learner Dec 7 '12 at 8:36
1  
Indeed, if $G$ is the group of symmetries of a regular hexagon, it has order $12$, and its center has order $2$, so the quotient has order $6$. If $G$ is the group of symmetries of a square, it has order $8$, and its center has order $2$, so the quotient has order $4$. Finding an example with quotient of order $25$ is a bit harder. –  Gerry Myerson Dec 12 '12 at 2:48
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