To find posible order of $\frac {G}{Z (G)} $ If $Z (G)$ denotes centre of group $G,$ the order of quotient group $\frac {G}{Z (G)} $
cannot be 
$(A) 4.$
$(B)6.$
$(C)15.$
$(D)25.$
I know that if $G/Z(G)$ is cyclic then $G$ is abelian.
 A: If $H=(G/Z(G))$
then say if $o(H)=15$
then $H$ will have  a unique  sylow 3 subgroup and a  unique sylow 5 subgroup ;hence by the internal direct property of groups $H\cong \mathbb Z_3\times \mathbb Z_5$ and $gcd(3,5)=1$
 will imply $H\cong \mathbb Z_{15}$
Thus $H$ is cyclic .If $G/Z(G)$ is cyclic then G is abelian and hence $G=Z(G)$ contradicting our assumption
Thus C will be never true
A: In the comments the OP states that he knows that "if $G/Z(G)$ is cyclic then $G$ must be abelian". This is the key idea. The secondary idea is: when is a group of order $pq$ or $p^2$ cyclic, for $p$ and $q$ primes?


*

*If $p$ and $q$ are non-equal then it is related to $p$ dividing
$q-1$, but I presume that this will have already been covered as it
requires a lot of work otherwise, so this would be an odd question!

*For the $p^2$ case then think about $\mathbb{Z_p}\times\mathbb{Z_p}$.


So, if $G/Z(G)$ is cyclic then what must is actually be? Can it be, for example, cyclic of order $7$? what about $21?$ No! What must it be?
Then go through the list working out which orders can only give you cyclic groups, using your prior knowledge about groups of order $pq$ and of order $p^2$.
As has been pointed out in the comments by @Tobias, you then need to prove that for each of the remaining three orders $x_1$, $x_2$ and $x_3$ there exists a group $G_i$ such that $G_i/Z(G_i)$ has order $x_i$. For example, there exists a non-abelian group of order $8$, so $G/Z(G)$ has order $4$ (why?).
