This question already has an answer here:

I want to show that if a finite abelian group has elements of order $m$ and $n$ then it will have an element of order $\text{lcm}(m,n)$.

First I proved the lemma if $a$ has order $m$ and $b$ has order $n$ with $m,n$ coprime, then $ab$ has order $mn$. This is because the subgroups $\langle a \rangle$ and $\langle b \rangle$ have trivial intersection (because the order of their intersection must divide their orders $m$ and $n$, which is a consequence of it being a subgroup of both), which implies $a^{i} \in \langle b \rangle$ iff $i \equiv 0 \pmod n$, and similarly $b^{i} \in \langle a \rangle$ iff $i \equiv 0 \pmod m$. Using that we deduce that $(ab)^i = a^i b^i = 1$ iff $i \equiv 0 \pmod {mn}$.

So if $a$ was an element of order $m$ and $b$ and element of $n$ with $g = \gcd(m,n) \not = 1$ I thought that $\text{lcm}(m,n) = \frac{m}{g}n$ so and $\frac{m}{g},n$ are coprime so I should construct an element (from $a$) with order $\frac{m}{g}$ then conclude the theorem by the lemma. For the construction I think it's just $a^g$.

I just have a nagging doubt about the correctness of the second proof, did I miss some important detail? Also if there are any neater ways to prove this (which don't depend on the structure theorem) I would like to learn them too.


marked as duplicate by Marc van Leeuwen, Chris Godsil, Ian Miller, kjetil b halvorsen, user147263 Dec 11 '15 at 1:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ Saying that $m/g$ and $n$ are coprime is wrong. Consider $n=12$, $m=18$, the standard example. What is true is that $m/g$ and $n/g$ are coprime. $\endgroup$ – Marc van Leeuwen Mar 28 '14 at 14:11

Everything you said seems fine. Using additive notation, you only need to consider the subgroup $sa + tb$ of $G$. This group contains the element $a+b$ which has the property

$$ \hbox{lcm}(m,n) \cdot (a + b) = 0 $$

so the order of $a+b$ is at most $\hbox{lcm}(m,n)$.

Moreover the order must be equal to $\hbox{lcm}(m,n)$, because any smaller order could be divisible by both $m$ and $n$.

  • $\begingroup$ You mean be divisible by both $m$ and $n$. And we must justify the lemma that the order of $a+b$ must be divisible by the orders of $a$ and $b$. :-) $\endgroup$ – anon Dec 11 '12 at 1:03
  • $\begingroup$ @anon thanks, fixed the mistake $\endgroup$ – orlandpm Dec 11 '12 at 1:08
  • $\begingroup$ No, everything is not fine. $\endgroup$ – Marc van Leeuwen Mar 28 '14 at 14:09
  • $\begingroup$ Also the argument given here is wrong, as you can see by taking $b=-a$, then the order of $a+b$ is $1$, which is not lcm$(n,n)$ (one does of course have $n=m$ in this case). $\endgroup$ – Marc van Leeuwen Mar 28 '14 at 14:16

Saying that $m/g$ and $n$ are coprime is wrong. Consider $n=12$, $m=18$, the standard example. What is true is that $m/g$ and $n/g$ are coprime. But if you want to complete the proof using the elements $a^g,b^g$, then you will realise that although these now have coprime orders, the least common multiple (=product) of those orders is no longer $\def\lcm{\operatorname{lcm}}\lcm(n,m)$, but $\lcm(n,m)/g$.

So instead another idea is needed. The best I have been able to think of is treat all prime numbers $p$ dividing $nm$ separately; if $p$ divides the order of only one of $a,b$ then there is nothing to do for $p$, but if it divides both orders, then keep the element in which the multiplicity is highest (choose one in case of a tie) and kill the factors $p$ in the order of the other element (say it was $b$) by replacing $b$ by $b^{p^i}$ (where $i$ is the minority multiplicity). This modification is designed to leave the least common multiple of the orders of $a,b$ intact, and when all primes have been processed the greatest common divisor has become$~1$. After these modifications, the element $ab$ has the desired order.