Prove any group of order three is abelian?

Question

Prove any group of order three is abelian ?

Answer 1

Let G be a group of order 3, so that G = {e, a, b}.



Then

ea = a = ae

eb = b = be



So the only thing we need to show is ab = ba. This will be done by showing ab = e, so that b = a^-1 and ba = e = ab.

Assume ab = a, then by left multiplication by a^-1, we get b = e, a contradiction.

Assume ab = b, then by right multiplication by b^-1, we get a = e, a contradiction.

Therefore ab = e, and it follows that ab = ba.



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For George P:

While what you say ultimately is all true, I'm wondering about your introduction of y as x^-1. A priori, we could have x = y, so that you were dealing with just two elements. Then, again a priori, we might have another element z = z^-1, giving a strange group of order 3.

(of course, this can't happen, since then xz must be some fourth element, but that hadn't been proven)

Answer 2

maximum difficulty-free evidence: the team must have order a million, 2, 3, or 4. Any team of precise order is cyclic, so as that seems after orders 2 and three. Any team of order a million has one aspect e, and positively e commutes with itself, because powers of any aspect holiday with one yet another. That leaves order 4. enable G be a collection of order 4, and enable x be an aspect of maximal order in G. If x has order 4, then G is cyclic, and we are carried out. So, because the order of an aspect divides the order of the team, x must have order a million or 2. on account that G has a non-identity aspect, and x has maximal order, we deduce that x does no longer have order a million. So, x has order 2. enable y be any aspect of G, except x or the identity aspect. We declare that G = {a million,x,y,xy}, the position a million is the identity aspect of G, and that more than a number of those aspects holiday with one yet another. actually, y isn't a million or x, and on account that x has order 2, and no aspect of G has better order, y^2 might want to be a million. also, xy can't be any of one million, x, or y, by using the existence of inverses. besides, on account that all aspects of G have order at maximum 2, the aspect xy has order 2, so as that a million=xyxy. yet x and y each and every have order 2, so if we multiply this equation by using yx, we get yx = yxxyxy = yyxy = xy. for this reason, G is abelian. QED For a 2d evidence, yet one you should no longer be allowed to apply, word the outcome that any team of order both p or p^2 is abelain. For a third evidence, use the actual undeniable reality that any cyclic team is abelian, and that the middle of a p-team is nontrivial. If the middle of G is Z, argue that G/Z has order 2 or a million, and is for this reason cyclic. Then, use the definition of "middle" to finish the evidence.

Answer 3

Let's think about the elements in a group of order 3. One of them must be the identity element e. Let's say x is another element in the group, not equal to e. By the group axioms, x must have an inverse element in the group y = x^-1. These are the 3 elements of the group.

Let's check the multiplication table for commutativity.

ex = x =xe (because e is identity)

ey = y = ye (again, because e is identity)

xy = e = yx (because y is x^-1).



I think that takes care of everything. It's an abelian group.