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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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    Sorry...The above argument is afterall wrong. I forgot to consider the action of $Z_m$ on $Z_n$.Apology here.2012-12-11

1 Answers 1

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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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    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.2012-12-12