The discussion centers around a group theory problem involving groups of even order and the existence of an element of order 2. The original poster attempts to demonstrate that in a group G of even order, there exists an element g, distinct from the identity, such that g squared equals the identity.
The discussion is ongoing, with participants examining the original argument and questioning its assumptions. Some guidance is provided regarding the properties of inverses in groups, but no consensus has been reached on the correctness of the original claim.
Participants note that the removal of the identity from a group of even order results in an odd number of elements, which raises questions about the pairing of elements with their inverses. The implications of this observation are under consideration.
I see your argument. Is this true though? Can't ab = 1, bc = 1, cd = 1 etc etc but ba not equal 1?HallsofIvy said:The identity is, of course, its own inverse. Since the group is "of even order", i.e. it has an even number of elements, removing the identity leaves an odd number of elements. Pairing each number with its inverse, what happens?
PsychonautQQ said:I see your argument. Is this true though? Can't ab = 1, bc = 1, cd = 1 etc etc but ba not equal 1?