The discussion revolves around a group theory problem where the original poster is tasked with demonstrating that a group \( G \) in which every non-identity element has order two is commutative. The context involves properties of group elements and their orders.
The discussion has seen some agreement on the initial approach, with participants affirming the correctness of the reasoning presented. However, there is also a desire for additional methods to demonstrate the same result, indicating an ongoing exploration of the topic.
Participants note that the order two aspect of the problem is distinct from a subsequent question regarding generators of \( \mathbb{Z}_n \) and their relationship to relative primality, suggesting a separation of the two topics in their discussions.
VeeEight said:Looks good to me.
e = abab implies
b = ababb = aba which implies
ba = abaa = ab
halvizo1031 said:Thanks! Is there another way i can show this?
Dick said:Why do you need another way? What did you have in mind?
halvizo1031 said:well i just don't want to have the same answer as someone else.
Dick said:It's a simple problem. There's a simple answer. There's a few different permutations on the expression of that answer, but they are all really the same. Wouldn't it be better to move on to the next problem?
halvizo1031 said:That's true. I'm having trouble showing that an element k is a generator of Zn if and only if k and n are relatively prime.
Dick said:The order 2 part of the problem has nothing to do with proving k is a generator of Zn if k and n are relatively prime. Try thinking about them independently.