Herstein's Famous Problem

**ThePerfectHacker** · November 1st 2007, 05:17 PM

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.

**ThePerfectHacker** · November 1st 2007, 05:52 PM

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.