G x H: Proving Cyclicity with GCD = 1

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Homework Help Overview

The problem involves proving that the direct product of two cyclic groups, G and H, is also cyclic when the orders of the groups are coprime (gcd(m,n) = 1). The original poster presents an attempt at a solution based on the properties of least common multiples and group orders.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to use the relationship between the orders of elements and the groups to argue that the direct product is cyclic. Some participants question the validity of assuming that any element's order equals the group's order, suggesting the need for careful selection of generators.

Discussion Status

Participants are exploring the implications of the hint provided in the problem statement regarding the generators of the groups. There is an acknowledgment that the proof requires ensuring that the chosen elements are indeed generators, and some guidance has been offered regarding this aspect.

Contextual Notes

There is a hint in the problem suggesting that G and H are generated by specific elements, which may influence the approach to the proof. The discussion highlights the need to consider the properties of cyclic groups and their elements carefully.

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Homework Statement


Let G and H be cyclic groups, with |G| = m and |H| = n. If gcd(m,n) =1, show that G x H is cyclic.


The Attempt at a Solution


Let a = (g,h) in G x H. Then |a| = lcm (|g|,|h|).
Since gcd(m,n)=1, then lcm (m,n) = mn.
Thus lcm (|g|,|h|) = lcm (m,n) = mn.
so <a> = G x H has mn elements and a cyclic group.
Right?
 
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Basically ok. But you want to be a little more careful with your choice of g and h. It's not true that for every element g in G that |g|=|G|. E.g. |e|=1. And G could have nontrivial subgroups. You'd better make sure that g and h are generators.
 
so.. you mean I have to make sense that for every element g in G that |g|=|G|?
If so..
Actually, the problem has Hint that [ G=<g> and H = <h>, show |(g,h)| = mn].
If consider this hint, my answer is ok?
or any other thing needs to prove?
 
Last edited:
The point is that at least one g in G satisfies |g|=|G|. Otherwise, would it be a cyclic group? That's what the proof is missing.
 

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