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Math Help - Herstein's Famous Problem

  1. #1
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    Herstein's Famous Problem

    There is this one problem in Herstein's book which recieved more comments than anything else.

    "Show that is a finite abelian group has subgroups of orders n and m then it has a subgroup of order \mbox{lcm}(n,m)".

    An unfair way to solve this problem is to use the fundamental theorem on finite abelian groups, meaning since n,m divide |G| then so does their least common multiple. But then there exists a subgroup of that order.

    I found a simplified proof to this using the first Sylow theorem. I will prove a stronger result. Let n be a divisor of the order of a finite abelian group then it has a subgroup of that order. This is more than what is being asked but it works "easily".*

    First, let H,K be subgroups of a finite group. So that \gcd (|H|,|K|)=1. Then H\cap K = \{ e \}, because if x\in H and x\in K then \mbox{ord}(x) is a divisor of |H| and |K| since their orders are relatively prime it means x=e. Now if H is such a group so that khk^{-1} \in H for all h\in H , \ k\in K then the group product HK = \{hk|h\in H, k\in K\} is a subgroup. In particular if the group is abelian then this property must hold and so the group product is always a subgroup. The fundamental observation is that if H,K are subgroups with orders n,m which are relatively prime then HK is a subgroup (we are assuming we are working in a abelian group of course) is a subgroup of order nm. The proof behind this rests in the identity that |HK|\cdot |H\cap K| = |H|\cdot |K|, since as we explained that |H\cap K|=1 it means |HK|=nm. Now finally we can complete the proof. Let d be a divisor of the order of the finite abelian group. Certainly if d=1 then there is a subgroup of this order. So it is safe to assume that d>1, in that case the we can factorize d=p_1^{a_1}...p_k^{a_k} where each p_i^{a_i} is a divisor of |G|. By Sylow's theorem there is a subgroup H_i having this order. If we let H be the subgroup product of all H_i then it has order d because \gcd (|H_i|,|H_j|)=1 for \delta_{ij}=0. Now by above it shows that H has order d.




    *)I use this expression because maybe Sylow's first theorem is considered heavy machinery. I do not consider it to be, but I have seen others think like that.
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  2. #2
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    I found a way to generalize Herstein's problem.

    Note in this proof we do not really need that fact the group is abelian. We just need the fact that all sub-group products are subgroups. As a result, let me give this group a name in honor of Grothendieck.

    Defintion: A "Grothendieck group" is a group such that if H,K are subgroups then HK is a subgroup.

    Note, Grothendieck groups do not need to be abelian. For instance, Dedekind groups are Grothendieck groups because all its subgroups are normal. So the quaternion group which is Dedekind but not abelian is Grothendieck.

    Here is the generalization:
    "If G is a finite Grothendieck group and has subgroups of orders n,m then it will have a subgroup of order \mbox{lcm}(n,m)".

    Not only does it generalize Herstein's problem it generalizes the problem of the converse of the theorem of Lagrange. By extending the converse to a larger class or groups.
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    Here is another side bar.
    Max Zorn of Zornís lemma had only one PhD student.
    Who was it? I.N. Herstein
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  4. #4
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    Quote Originally Posted by Plato View Post
    Here is another side bar.
    Max Zorn of Zornís lemma had only one PhD student.
    Who was it? I.N. Herstein
    That is a cool fact. I just checked his math geneology. I also found that Max Zorn was a student, the only PhD student, of Emil Artin.
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