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I am looking for a group that has exactly three maximal abelian subgroups.

I thought about the quaternion group. $G=Q_8 = \langle x,y \mid x^4=1, x^2=y^2, yxy^{-1}=x^{-1}\rangle$.

$Z(G) = \mathbb{Z}/2$ and one of the abelian subgroups is $\langle x\rangle = \{e,x,x^2,x^3\}$.

But I have problems to find the other ones. I don't really know how to construct them.

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    Well there must be a maximal subgroup containing $y$? Can you think of one. And there must be one containing $xy$.2012-11-23

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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 $Q$.