3 ms·
Note on the problem that a group G of even order contains an element of order 2. Partition G into classes {{x,x^{-1}|x\in G}. This is a partition since inverses
by Xlp-Thlplylp 18y ago
Note on the problem that a group G of even order contains an element of order 2. Partition G into classes {{x,x^{-1}|x\in G}. This is a partition since inverses are unique. For x\in G, call {x, x^{-1}} the class of x. At least one other element x besides the identity has a class of size one. Otherwise, the order of G would be 1 + 2*(# of classes of size 2) which is odd. Hence there is an x with x != 1 and x = x^{-1}; i.e., an element of order 2.
- technoguyrob 18y agoThanks, that's exactly the kind of comments I was looking for. That was exactly what I thought in my head, but it came out very distorted and ugly on paper. I'll update the solution later. Do you want credit?
- Xlp-Thlplylp 18y agoCredit for this is optional. Another way to see obtain a partition of G is to observe that the relation x ~ y if and only if x = y or x = y^{-1} is an equivalence relation on G. The equivalence classes induce the partition above. The point is that any two equivalence classes {x, x^{-1}},{y,y^{-1}} are either disjoint or equal, and their union is G. But this is obvious (the word 'obvious' means "I thought of it.")