Quotient group is cyclic or not? Let G be the quaternion group and Z(G) is center of the group G then the quotient group G/Z(G) is isomorphic to-
A) G
B) ({1, -1}, .)
C) Klein's 4-group
D) Group of integers modulo 4


*

*I've figured out that option A and B are wrong. Given quotient group is abelian group of order 4.


-Group in Option A) isn't abelian.
-Group in option B) is of order 2.
Now to choose right one from C and D, i need to know that given quotient group is cyclic or not? But  How to do that?
 A: You can compute the center and you will get that it is a group of order $2$ such that the quotient has order $4$ as you mentioned. Therefore C) or D) could be correct by now. If you are considering the elements in the quotient you will see that every non-trivial element has order $2$ and therefore you will get the Klein-$4$-group. So yes, check whether the quotient is cyclic by computing the orders.
Let me give an example. The order of $i \cdot \lbrace 1,-1 \rbrace \in Q_8/Z(Q_8)$ is $2$, as $i \not\in \lbrace 1,-1 \rbrace$ and
$(i \cdot \lbrace 1,-1 \rbrace)^2 = i^2 \cdot \lbrace 1,-1 \rbrace = -1 \cdot\lbrace 1,-1 \rbrace = 1 \cdot \lbrace 1,-1 \rbrace$.
Now do the computation for the other $2$ non-trivial elements of the quotient.
A: One way to do this is to consider the presentation of the group. Note that 
$$Q_8 = \langle i,j,k~|~i^2=j^2=k^2=ijk=-1\rangle.$$
Since $Z(W_8) = \{1,-1\}$, in $Q_8/Z(Q_8)$ we have that $ijk = 1$, or $ij = k$. Thus, we know that $(ij)^2 = 1$ in the quotient, implying that $ij=ji$. Thus, we have a presentation
$$Q_8/Z(Q_8) = \langle i,j~|~i^2 = j^2 = 1,~ij=ji\rangle.$$
This is exactly the presentation of the Klein four group.
