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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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Why have you thought about the quaternion group? Do you have some reason to believe it will work? –  Chris Eagle Nov 23 '12 at 19:07
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Well first I thought it might be some nice non-abelian small group. like $D_{4}$ or $Q_{8}$, and then, to be honest, I looked in the internet what they said about these groups, and I found an article where they said the $Q_{8}$ had 3 abelian subgroups. So I am looking for them ;) –  Kathrin Nov 23 '12 at 19:24
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Well there must be a maximal subgroup containing $y$? Can you think of one. And there must be one containing $xy$. –  Derek Holt Nov 23 '12 at 20:09

1 Answer 1

up vote 1 down vote accepted

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. $<i>,<j>$ and$<k>$. 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 $Q$.

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