How to prove existence of an element of order $\text{lcm} (m,n)$ in an abelian group having elements of order $m$ and $n$?

Asked Nov 18 '20 at 07:44

Active Nov 18 '20 at 08:57

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This particular question was asked in my algebra quiz and I was unable to think of what result I should use to prove this statement.

Let $G$ be a finite abelian group and $a,b \in G$ with $|a|=m$ and $|b|=n$. Then prove that there always exists an element of $G$ whose order $= \text{lcm}(m,n)$

I proved that there always exists an element whose order divides $\text{lcm} (m,n)$. If $\gcd (m,n) = 1$ then $|ab|=\text{lcm}(m,n)$ .But how to prove it in case when $\gcd(m,n)>1$?

Which result should I use?

edited Nov 18 '20 at 08:57

ArsenBerk

13,388

asked Nov 18 '20 at 07:44

Hint: if $g=\gcd(m,n)$, consider the orders of $a^g$ and $b^g$, use the fact you've already established to find the order of $a^gb^g$, and then draw conclusions about the order of $ab$. – Greg Martin Nov 18 '20 at 07:49
https://math.stackexchange.com/questions/3737548/let-g-be-an-abelian-group-with-elements-x-y-of-orders-m-and-n-respectiv?rq=1 – Sep 20 '21 at 12:02

How to prove existence of an element of order $\text{lcm} (m,n)$ in an abelian group having elements of order $m$ and $n$?

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