On order of subgroups in abelian groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T14:26:37Z http://mathoverflow.net/feeds/question/19092 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19092/on-order-of-subgroups-in-abelian-groups On order of subgroups in abelian groups J. H. S. 2010-03-23T06:49:14Z 2012-08-07T04:01:55Z <p>I wonder whether any of you guys has already read the homonymous note by R. Beals in the December 2009 issue of the <strong>Monthly</strong>.</p> <p>If so, <strong>would you be so kind as to let me know about the main ideas in Beal's approach</strong>? As you know, the whole point of his note is to present a solution to the following exercise in Herstein's <em>Topics in Algebra</em>:</p> <p>Let <strong>G</strong> be an abelian group having subgroups of order <em>m</em> and <em>n</em>. Prove that <strong>G</strong> also possesses a subgroup of order <strong>lcm</strong>(<em>m</em>, <em>n</em>).</p> <p>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 <em>authorized</em> 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.".</p> <p>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 <strong>gcd</strong>(<em>m</em>, <em>n</em>)=1?</p> <p>Thanks in advance for you insightful replies.</p> <p>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. </p> http://mathoverflow.net/questions/19092/on-order-of-subgroups-in-abelian-groups/19096#19096 Answer by Robin Chapman for On order of subgroups in abelian groups Robin Chapman 2010-03-23T07:44:17Z 2010-03-23T07:44:17Z <p>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 $\mathrm{lcm}(m,n)$.</p> <p>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.</p> <p>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.</p>