Abelian Group Proof: Example to Show Necessity

[kathrynag]

Homework Statement

If G is an abelian group, then (ab)^2=a^2b^2 for all a, b in G. Give an example to show that abelian is necessary in the statement of the theorem.

Homework Equations

The Attempt at a Solution

Abelian implies commutativity.
a*b=?b*a
a^2b^2=b^2a^2
For example ab=ba must be true for the statement to work.

 

[waht]

take for example a^2 b^2 and expand it to aabb

what can you do if ab is abelian?

 

[kathrynag]

aabb
Implies commutativity so aabb=bbaa
I think there's something with inverse to make this work

 

[waht]

a(ab)b

 

[kathrynag]

Ok so a(ab)b and b(ab)a
so this implies ab=ba

 

[waht]

Ok so a(ab)b and b(ab)a
so this implies ab=ba

the definition of abelian is not axb = bxa so can't do that

so instead we can swap the ab

 

[kathrynag]

I don't really understand. You're saying we can swap the ab, but I don't gte what you mean by this.
(ab)ab=(ab)ba

 

[matt grime]

I don't know what what it making you do. The point was to show that you can find G non-abelian where (ab)^2 is not equal to a^2b^2 for some a and b.

 

[kathrynag]

Yeah and I wanted to prove that abelian implied (ab)^2=a^2b^2
Is there a hint for an a and b to use where the 2 aren't equal?

 

[Count Iblis]

(ab)^2 = a^2 b^2 --->

abab = aabb -------->

cancel the leftmost a and the rightmost b:

ba = ab

 

[waht]

well

a^2b^2 = aabb = a(ab)b

because they are abelian ab = ba

so aabb = a(ba)b = (ab)(ab)

 

[Count Iblis]

well

a^2b^2 = aabb = a(ab)b

because they are abelian ab = ba

so aabb = a(ba)b = (ab)(ab)

But you are supposed to prove that they are Abelian.

 

[kathrynag]

Ok I understnd how to prove it now.
I'm having trouble with the example to show abelain is necessary.

 

[Count Iblis]

Ok I understnd how to prove it now.
I'm having trouble with the example to show abelain is necessary.

It is necessary because:

(ab)^2 = a^2 b^2 --->

abab = aabb -------->

cancel the leftmost a and the rightmost b:

ba = ab

So, we have that

(ab)^2 = a^2 b^2 implies that ab = ba

This is the same as saying:

Not[ab = ba] implies Not[(ab)^2 = a^2 b^2]

Or, in plain English, if you don't have the Abelian property, you don't have that (ab)^2 = a^2 b^2. So, the Abelian property is a necessary assumption for the identity to hold.

 

[kathrynag]

Hmmm, I think I might see this better if I saw a way where it doesn't work. For example (ab)^2 not equaling a^2b^2. Therefore, not abelian

 

[Count Iblis]

Hmmm, I think I might see this better if I saw a way where it doesn't work. For example (ab)^2 not equaling a^2b^2. Therefore, not abelian

But that's equivalent to Abelian ---> (ab)^2 =a^2b^2

which is the trivial case.

If you know that this is true (so Abelian implies the identity), then should you encounter a case where the identity is not satisfied, you know that the group cannot be Abelian.

 

[kathrynag]

that's what I'm trying to do to find a case now where the identity is not satisfied to show then that the group is not abelian

 

[matt grime]

Um, as count Iblis has pointed out *any* non-abelian group will provide a counter example.

 

[Count Iblis]

that's what I'm trying to do to find a case now where the identity is not satisfied to show then that the group is not abelian

You can take some group of matrices. Matrix multiplication does not commute. So, it should be easy to find a particular example where the identity does not hold.

But such a particular example does not show that the fact that the identity does not hold implies that the group is non-abelian. To show that, you need to prove that whenever the identity does not hold, you also have that the group is not abelian.

 

[kathrynag]

Oh, that makes sense.