Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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
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
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 $Q$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.