Suppose $G$ is a finite abelian group and has two element $a$ and $b$, such that $\circ(a)=m$ and $\circ(b)=n$ and $lcm(m,n)\neq m,n$. Is it true that $\circ(ab)=lcm(m,n)$? Thanks in advance.
Asked
Active
Viewed 655 times
7
-
What did you try? – egreg Nov 12 '13 at 10:19
-
I have try many things, but none of them work. Any idea? – bor Nov 12 '13 at 10:20
-
1I have shown $\frac{mn}{gcd(m,n)^2} \mid \circ(ab)$. – bor Nov 12 '13 at 10:21
-
I have also shown that $G$ has an element of order $lcm(m,n)$. But could not show that it is $ab$. – bor Nov 12 '13 at 10:22
2 Answers
6
The statement is not true. In $\mathbb{Z}/30 \mathbb{Z}$, $o(3) = 10$ and $o(5) = 6$, but lcm$(6,10) = 30$ and $o(3+5) = o(8) = 15$.
Arthur
- 5,524
5
But it is true that if $\operatorname{ord}(a)=m$, $\operatorname{ord}(b)=n$ and $\operatorname{gcd}(m, n)=1$ then $\operatorname{ord}(ab)=mn=\operatorname{lcm}(m, n)$.
In general, one can say the following:
Proposition. If $G$ is a finite abelian group, and $\operatorname{ord}(a)=m$, and $\operatorname{ord}(b)=n$. Then $$\frac{mn}{\gcd(m, n)^2}\mid \operatorname{ord}(ab) \mid \frac{mn}{\gcd(m, n)}=\operatorname{lcm}(m, n)$$
It looks like you have shown $\frac{mn}{\gcd(m, n)^2}\mid \operatorname{ord}(ab)$ as you indicated in the comments. Very nice!
For the proof of the above fact, see the article
Dieter Jungnickel, On the Order of a Product in a Finite Abelian Group. Mathematics Magazine, Vol. 69, No. 1 (Feb., 1996), pp. 53-57
Prism
- 11,162
-
3Apparently the PDF is freely available at http://www.maa.org/sites/default/files/Dieter_Jungnickel20395.pdf. – lhf Nov 12 '13 at 11:32
-
Nice find! I was using my university's access to JSTOR, so it didn't occur to me that the article could be found elsewhere freely. Thanks a lot! – Prism Nov 12 '13 at 11:37