Introduction to Group Theory - Abstract Algebra

[descendency]

Homework Statement

Prove that if (ab)² = a²b² in a group G, then ab = ba.

Homework Equations

* For each element a in G, there is an element b in G (called the inverse of a) such that ab = ba = e (the identity).
* For each element in G, there is a unique element b in G such that ab = ba = e.
* The operation (under which the group is defined) is associative; that is, (ab)c = a(bc).

The Attempt at a Solution

(ab)^{2} = (a)^{2}(b)^{2}
(a)^{-1}(ab)^{2}(b)^{-1} = (a)^{-1}(a)^{2}(b)^{2}(b)^{-1}
(a)^{-1}abab(b)^{-1} = (a)^{-1}aabb(b)^{-1}
ebae = eabe
ba = ab.

Is this allowed? (Keep in mind, the operation which the group is defined over is not necessarily multiplication. it may be composition.)

 

[NoMoreExams]

Sure is but your pre last line should be ba = ab I believe.

Also your 2nd "given" is the same as your first, you probably meant something else?

 

[descendency]

Sure is but your pre last line should be ba = ab I believe.

Also your 2nd "given" is the same as your first, you probably meant something else?

The second given is the uniqueness statement. The first given is the rule that an element in a group has at least an inverse (not necessarily unique). I said "the" when I meant "an".

edit: Thanks for correcting my mistake. It's a typo. I had it right on paper, I just didn't type out what I wrote down for some reason. . .

 

[NoMoreExams]

Then that seems redundant, line 2 is stronger than line 1.

 

[Unco]

The Attempt at a Solution

(ab)^{2} = (a)^{2}(b)^{2}
(a)^{-1}(ab)^{2}(b)^{-1} = (a)^{-1}(a)^{2}(b)^{2}(b)^{-1}
(a)^{-1}abab(b)^{-1} = (a)^{-1}aabb(b)^{-1}
ebae = eabe
ba = ab.
Is this allowed?

Your work is fine, where I read it as (line 1) => (line 2) => ... .

However, depending on the level of pendency of your instructor you may be required to justify your use of associativity. E.g., the left-hand side is (ab)^2 = (ab)(ab) = a(b(ab)) = a((ba)b), and the right-hand side is a^2b^2 = (aa)(bb) = a(a(bb)) = a((ab)b); then multiply both sides on the left by a^{-1}, and so on.