I am working through Rotman's abstract algebra and the above question has been giving me problems. I have a solution book for it which gives the following:
We know $Z(Q) = \{\pm E\}$ so $Q(Z)$ has order $4$. We also know that for all $M \neq \pm I$, $M^2 = I$. Thus every nonidentity element of $Q/Z(Q)$ has order $2$ and hence $Q/Z(Q) \cong V$.
My question is how does the last step follow? Every element having order $2$ means it's abelian but even if all elements have the same order in $V$ and $Q/Z(Q)$ it does not mean that they are isomorphic. So how is proving the order sufficient?
By Lagrange's Theorem,
$$\begin{align} |Q/Z(Q)|&=|Q|/|Z(Q)|\\ &=8/2\\ &=4. \end{align}$$
As noted in the comments, there is only two groups of order four up to isomorphism: the Klein four group $V$ and $\Bbb Z_4$. But the latter has an element of order four; therefore, $Q/Z(Q)\cong V.$