- #1
clkt
- 6
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How do I proof that groups of an even order must have an element of order 2? I have a vague idea, but I don't know how to put my idea together.
Aside from identity, there are an odd number of elements in my group. So one element will not have a partner and will have to be multiplied by itself to cancel out. That element must have an order of 2 such that its square = identity. But how can I create the scenario where all elements have to pair up and cancel out? Thanks in advance.
Aside from identity, there are an odd number of elements in my group. So one element will not have a partner and will have to be multiplied by itself to cancel out. That element must have an order of 2 such that its square = identity. But how can I create the scenario where all elements have to pair up and cancel out? Thanks in advance.