It might be easier to think of the quaternion group as $Q = <1,i,j,k |i^2=j^2=k^2=-1,ijk=-1>$. It is not very hard to find all proper subgroups of $Q$. Note that any subset containing two elements out of $\{i,j,k\}$ immediately generates the entire group, since multiplication of these elements gives the third element and the square of either element gives $-1$, which commutes with all elements. Hence this subset generates $Q$. That leaves that any proper subgroup can contain either no elements or one element from $\{i,j,k\}$. If it contains no elements, the only subgroups are the trivial subgroup and $\{1,-1\}$. The other possibility gives you the subgroups generated by one element from the set, i.e. $,$ and$$. Check that these are abelian and the note the subgroup $\{1,-1\}$ is contained in either of these subgroups, but none of them is contained in either of the other two. Hence they are maximal abelian subgroups and the only ones contained in $Q$.