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Finite-Index Subgroups of Infinite Groups.

  1. Aug 3, 2011 #1
    Hi, All:

    Please forgive my ignorance here: let G be an infinite group, and let H be

    a subgroup of G of finite index . Does H necessarily have torsion? I can

    see if , e.g., G was Abelian with G=Z^n (+) Z/m , then , say, would have

    subgroups of finite index, but I can't tell if this is an iff condition.

    Any Ideas?

  2. jcsd
  3. Aug 3, 2011 #2
    I'm a bit confused. What does torsion mean for a group if it's not abelian? Without mofidifiers, torsion refers to torsion as a Z-module, which would force G to be abelian.

    In addition, any subgroup of a free abelian group has no torsion. This doesn't have anything to do with its index. For instance, 2Z in Z has finite index and no torsion. Do you mean G/H?
  4. Aug 4, 2011 #3
    Yes, Zhentil; sorry for my carelessness. I was considering first the case of Abelian groups, (using the standard decomposition as Z-modules) and then wondering if the results extend; and , yes, I was considering G/H.
  5. Aug 4, 2011 #4
    In that case, the answer is yes. A finite abelian group is a torsion module.
  6. Aug 5, 2011 #5
    I missed the other point, about whether it extends. You have to specify what you mean, but in general the answer is no. Consider Z2 + Z2, with Z2 as a subgroup, considered as a Z2 module. Then H has finite index, but both G and G/H are free.
  7. Aug 5, 2011 #6
    Hi, Zhentil; you're right; every subgroup of a free group is itself free (I think this is called Schreier's theorem or something). I am just wondering if there are known results whereby an infinite group , whether Abelian
    or not, has subgroups of finite index.

    As you pointed out, this cannot happen for
    free groups, by the above result (Schreier); but I am not sure whether I understood your result for general Abelian groups: let A be an infinite Abelian group (not free, of course) and let B be a subgroup of A. Under what conditions do we find B having finite index in A? I know if B has index 2 in A then B is normal in A , but I don't know if the converse is true.

    Also: how about for infinite non-Abelian groups A ? What results are there for A having subgroups B so that B has finite index in A? How about some examples?

    I wonder too, if A being infinite but A having a finite generating set would change the answer.

    Last edited: Aug 5, 2011
  8. Aug 6, 2011 #7
    The converse to the index two criterion does not hold. In general, it's highly unlikely that any structure theorems of the sort you mentioned are true. What I mean is that you can find nontrivial examples of groups where all nontrivial subgroups have finite index (Z), or groups where all proper subgroups have infinite index. I don't know of constraints on the index of a subgroup other than simplicity of the group ruling out index two subgroups.
  9. Aug 6, 2011 #8
    Whether the group is abelian does not matter. Take any group, cross with the integers, and you get subgroups of any index. You could ask tougher questions of course, like making your group have trivial center, or asking about possible indices of nonnormal subgroups, but I tend to believe that in this generality, absolutely anything is possible.
  10. Aug 6, 2011 #9
    For the other question, there's an easy way to know if a subgroup of an abelian group is normal. :) for the other question, if G is finitely generated and infinite, any index is possible. If it's finite, the index dividing the order is necessary and sufficient. If G is not finitely generated, knock yourself out. Anything is possible.
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