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Prove that if a abelian group has elements of order $m$ and $n$ then it has a subgroup of order equals to $lcm[m,n]$.

I am new to group theory so please explain....

Satvik Mashkaria
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  • Since G is abelian, you know any prime p, a factor of |G| is the order of N, a subgroup of G. That subgroup is a normal to G by G abelian. Now replace G by G/N. Then proceed to construct this group of order lcm(m,n) in this manner. – user45765 Aug 10 '14 at 16:51
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    I think this question is not a duplicate of the linked question. True, one can deduce the result here from that answer, but this one can be done with less machinery, and OP mentioned being new to group theory. – André Nicolas Aug 10 '14 at 17:07
  • @Andre Surely it would be better to just give an answer to the original question? The questions are identical, if you do not include the OPs efforts. – user1729 Aug 10 '14 at 18:34
  • @user1729: Our definitions of identical are not identical. This question is about elements of orders $m$ and $n$. The other is about subgroups. One can use the subgroups one to prove the elements one. But a proof of the elements one can be made more concrete. – André Nicolas Aug 10 '14 at 18:46
  • @Andre Ah, sorry, I missed the element/subgroup subtlety. However, this question is so standard that it has to be a duplicate, and a quick search reveals the following essentially identical question: http://math.stackexchange.com/q/255894/10513 (although I am sure a better example exists). – user1729 Aug 10 '14 at 19:28
  • I expected that the question was a duplicate of something. That was why I did not write out a construction (product for $m$ and $n$ relatively prime, take appropriate powers first otherwise to "separate" the factorizations). A closure as duplicate should lead OP and others to an answer that is directly relevant. – André Nicolas Aug 10 '14 at 19:36

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