A Proof Question?

Question

let G be the Abelian group.
Prove that G is abelian group iff (ab)^2 = a^2*b^2, where a,b belongs to G.

Answer 1

I guess if you want to show that G is an abelian group, your initial statement should be something like "let G be a group", not let G be the ableian group. Also I believe your second statement shoulld read "for all a,b in G" somewhere. Now assuming these, the proof is as follows. A group is abelian if ab=ba for all a and b. If that holds then we have
(ab)^2=(ab)(ab)=(ab)(ba)
=ab^2a=a^2b^2.
Conversely if (ab)^2=a^2b^2 then
(ab)^2=abab=a^2b^2, multiplying by a^{-1} on the left and b^{-1} on the right
a^{-1}(abab)b^{-1}=
a^{-1}(a^2b^2)b^{-1}, so
ba=ab.

Answer 2

To start the proof, G should only be assumed to be a "Group", not an Abelian group.

We assume. A Group G, and that (a*b)^2 = a^2 * b^2 for all elements a and b in G. For clarity, this relation can be rewritten as a*b*a*b = a*a*b*b. Since every element in G has an inverse under *, we can * both sides by a', the inverse of a. Thus, a' * a*b*a*b = a' * a*a*b*b. And by the Group definition, a' * a = e, the identity element. Thus e*b*a*b = e*a*b*b. And, by the definition of the identity element, e*b = b and e*a = a, and your relation is b*a*b = a*b*b

Repeat these steps, multiplying the right side of each side of the equation by b'. It must be the RIGHT side of EACH side, otherwise you are assuming commutativity! At the end, you should have b*a = a*b. Since b*a = a*b, then G is commutative, or Abelian, QED.

[Now, step back, and think. This is the point of the exercise. You have some group G, and it's associative, closed, with an identity and inverses. This particular group has an additional property (ab)^2 = a^2 * b^2. Using only the definition of a group, you can show that this property implies b*a = a*b.]

To finish this thought process, you can do one of two things: 1) verify that each step I use is a completely reversible step (an 'iff', not merely an 'if'), or 2) Start with a group G, and show that if b*a = a*b, then (ab)^2 = a^2 * b^2. I prefer #1, your professor may prefer #2.

Don't worry, it gets worse. I'm there for ya.

Answer 3

Let (G,*) be an Abelian group. Then, for all a,b ε G
a*b = b*a (by definition) so that
(a*b)^2 = a*b*a*b = a*a*b*b = a^2*b^2


Doug

Answer 4

HOMEWORK.
But if you can't do this one, you're in trouble anyway.

The previous poster showed one direction. For the other assume
(ab)^2=a^2 b^2.
Write this out:
abab=aabb.
Now cancel an a on the left and a b on the right to get
ba=ab.

Answer 5

(ab)^2=a^2*b^2 This is elementary and needs no proof what ever be the group