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You can still read through my solutions and point out all my errors if you like, though. ;) I've already had a couple classes, but I've never felt intimately c

You can still read through my solutions and point out all my errors if you like, though. ;) I've already had a couple classes, but I've never felt intimately comfortable with it. With several programming languages, I don't even need the occasional googling, nearly all the libraries (and of course basic syntax) are already in my head. When doing commutative algebra and Noetherian rings, I found myself looking back at ring theory for various properties about rings. Same goes for homological algebra: I had to keep going back to properties of modules. I understand it in more in terms of theorems and facts rather than mathematical intuition and maturity, and my goal is to change that around.

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 ago
  Thanks, 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 ago
    Credit 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.")

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 ago
Thanks, 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 ago
  Credit 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.")