I wonder whether any of you guys has already read the homonymous note by R. Beals in the December 2009 issue of the Monthly.

If so, would you be so kind as to let me know about the main ideas in Beal's approach? As you know, the whole point of his note is to present a solution to the following exercise in Herstein's Topics in Algebra:

Let G be an abelian group having subgroups of order m and n. Prove that G also possesses a subgroup of order lcm(m, n).

The funny thing about this proposal is that in subsequent editions of his book, Prof. Yitz would proclaim that he himself didn't have a solution using the authorized tools. Besides, he even went on to saying: "I've had more correspondence about this problem than about any other point in the whole book.".

Being aware of some of the history behind this little pearl, I'd really like to know what it is that Beals came up with. Is his approach crystal-clear? Is it somehow related to the standard attack of proving it first for the case gcd(m, n)=1?

Thanks in advance for you insightful replies.

P.S. The local library is the only access that I have to the literature. Unfortunately, they don't subscribe to any of the MAA periodicals.

1 Beals, Robert. "On Orders of Subgroups in Abelian Groups: An Elementary Solution of an Exercise of Herstein." The American Mathematical Monthly, Vol. 116, No. 10 (Dec., 2009), pp. 923-926; https://maa.tandfonline.com/doi/abs/10.4169/000298909X477032 https://www.jstor.org/stable/40391251

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    $\begingroup$ I am not completely familiar with what theorems are allowed for this, but proving that an abelian group has a subgroup of any order dividing the order of the group requires only very little. Namely the existence of a subgroup of order $p$ for a prime $p$ dividing the order of the group (a very simple proof for abelian groups) and the correspondence of subgroups containing $H$ and subgroups of $G/H$. Now the result follows by induction. $\endgroup$ – Tobias Kildetoft Mar 23 '10 at 11:53
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    $\begingroup$ I don't have immediate access to the article, but: why "homonymous"? $\endgroup$ – Pete L. Clark Mar 23 '10 at 13:52
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    $\begingroup$ I like Tobias's solution: it seems simple, direct and powerful. Since the Monthly article exists, I gather the use of quotient groups must be out of bounds? $\endgroup$ – Pete L. Clark Mar 23 '10 at 14:01
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    $\begingroup$ Actually, I meant same name as the present MO discussion. $\endgroup$ – José Hdz. Stgo. Mar 24 '10 at 1:58
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    $\begingroup$ I thought it was a reference to the painter, Homonymous Bosch. $\endgroup$ – Gerry Myerson Aug 1 '18 at 13:08

The question is to prove that if $H$ and $K$ are subgroups of a finite Abelian group or orders $m$ and $n$ then $G$ has a subgroup of order $\mathrm{lcm}(m,n)$.

Beals starts by doing the case where $H$ and $K$ are cyclic. He proves that $H$ is an internal direct product of cyclic groups of prime power orders. Then he proves that a product of cyclic subgroups of coprime orders is cyclic of the right order. The cyclic case is proved by breaking up $H$ and $K$ as products of cyclic prime power groups, taking the larger one for each prime and multiplying them up.

The general case follows roughly the same line. Proving that a product of subgroups of coprime orders has the right order is straightforward. But decomposing a subgroup into prime power factors using results earlier in Herstein is more involved. Beals uses Theorem 2.5.1 in Herstein that $|HK| = ~|H||K|/|H\cap K|$. Then Beals finishes the proof in the same way as the cyclic case.

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