Proved that a group is abelian

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In summary, the conversation discusses how to prove a group G is Abelian if and only if (ab)^-1 =a^-1*b^-1 for all a and b in G. The conversation includes a proposed proof using direct proof and assumptions, but it is not clear and does not explicitly mention what it means for a group to be Abelian. The importance of explicitly stating assumptions and using clear language in mathematical proofs is emphasized.
  • #1
Benzoate
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Homework Statement



Prove that a group G is Abelian if and only if (ab)^-1 =a^-1*b^-1 for all a and b in G.

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The Attempt at a Solution



Suppose (ab)^-1 =a^-1*b^-1 . Let (ab)^-1 *e = a^-1*e *b^-1*e. The multiply (ba)^-1 on the left side of the equation and b^-1 *a^-1 on the right side of the equation.
then (ab)^-1=(ba)^-1 and a^-1 *b^-1 =b^-1 *a^-1

then ((ab)^-1 *(ba)^-1)*e= (a^1*b^-1 *e)*(b^-1*a^-1*e
)

therefore, by method of direct proof, (ab)^-1 =a^-1 *b^-1
 
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  • #2
I think you just showed that (ab)^-1 =a^-1*b^-1 implies (ab)^-1 =a^-1*b^-1.

At no point in the proof do you even mention what Abelian means. That is surely a warning sign, isn't it?
 
  • #3
matt grime said:
I think you just showed that (ab)^-1 =a^-1*b^-1 implies (ab)^-1 =a^-1*b^-1.

At no point in the proof do you even mention what Abelian means. That is surely a warning sign, isn't it?

Other than failing to mention that Abelian means a*b=b*a , what else is wrong with my proof
 
  • #4
I'm a bit confused, too. Tip: Do the two steps necessary for showing the equivalence explicitely seperately, i.e.
1) Relation => Group is abelian
2) Group is abelian => Relation holds.
Using that [tex] \left( ab \right)^{-1} = b^{-1}a^{-1} [/tex] (make sure you understand why) that's two one-liners.
 
  • #5
Benzoate said:
Other than failing to mention that Abelian means a*b=b*a , what else is wrong with my proof

The simple fact that I can't decipher what it is that you tried to do, or what you think you did.
 
  • #6
I think I managed to understand it:
Benzoate said:
Suppose (ab)^-1 =a^-1*b^-1 . Let (ab)^-1 *e = a^-1*e *b^-1*e.
- Assume the relation holds (2nd version is just a rewrite with inserted identities).
The multiply (ba)^-1 on the left side of the equation and b^-1 *a^-1 on the right side of the equation. Then (ab)^-1=(ba)^-1 and a^-1 *b^-1 =b^-1 *a^-1
- Furthermore, assume the group is abelian.
then ((ab)^-1 *(ba)^-1)*e= (a^1*b^-1 *e)*(b^-1*a^-1*e). Therefore, by method of direct proof, (ab)^-1 =a^-1 *b^-1
- Then the relation holds.

Benzoate. If I understand that correctly, then from my translation you should be able to see the problem with your proof: You only say that the two statements do not contradict each other (not even completely sure if it sais that, but let's assume it did) but don't say that they are equivalent (which is a much stronger statement). From your statement, it could as well be that the only group for which both statements are true is the trivial group (with only one element).
 
  • #7
No, you cannot "assume the group is abelian" when you are trying to prove it is abelian!

Benzoate, it is true (and easy to prove) in any group that
(ab)-1= b-1a-1. You are given that (ab)-1= a-1b-1.

You should be able to combine those to show that ab= ba.
 
  • #8
Timo said:
I think I managed to understand it:

- Assume the relation holds (2nd version is just a rewrite with inserted identities).

- Furthermore, assume the group is abelian.

- Then the relation holds.

Benzoate. If I understand that correctly, then from my translation you should be able to see the problem with your proof: You only say that the two statements do not contradict each other (not even completely sure if it sais that, but let's assume it did) but don't say that they are equivalent (which is a much stronger statement). From your statement, it could as well be that the only group for which both statements are true is the trivial group (with only one element).

What if I were to supposed that (ab)^-1= b^-1 *a^-1. then I can say that if (ab)^-1 =b^-1*a^-1 , then b^-1 * a^-1 =(ba)^-1 => (ab)^-1 =(ba)^-1 also => b^-1*a^-1 =a^-1 *b^-1

Therefore , (ab)^-1 = a^-1 *b^-1
 
  • #9
Benzoate, can you show how you get each step... You should show that:

(ab)^-1=a^-1*b^-1 => ab = ba

as the first proof... so start with the assumption that (ab)^-1=a^-1*b^-1, and show that this leads to ab = ba. I think you almost did this but I found it a little hard to follow.

and then for the second proof you should show that:

ab = ba => (ab)^-1=a^-1*b^-1
 
  • #10
Benzoate said:
What if I were to supposed that (ab)^-1= b^-1 *a^-1.

Why would you suppose that? It is clearly true: just compose (ab) with b^-1a^-1.
then I can say that if (ab)^-1 =b^-1*a^-1 , then b^-1 * a^-1 =(ba)^-1 => (ab)^-1 =(ba)^-1 also => b^-1*a^-1 =a^-1 *b^-1

Therefore , (ab)^-1 = a^-1 *b^-1

You see, I'm still lost as to what it is that you think you have proved.

Suppose G is abelian, show that this implies (ab)^{-1}= a^-1b^-1.

Now, say where you assume abelianness, and use words, like 'hence we have shown that G abelian implies...' so that people can understand what you have written and why you have written it and what you think you have shown.
Once you've done that, reverse the process, and show that (ab)^-1 = a^-1b^-1 for all a,b imples G is abelian. Again, use words, not just a string of a,b,^,-1,=,> symbols so that we follow what it is you think you're doing.

Anyone can string together abstract symbols - you should be trying to show that you understand what stringing them together does and why. Do good textbooks look like this? Do papers? No, they contain lots of words and explanations. For a reason: maths is hard to comprehend at the best of times, but next to impossible if you don't write clearly.
 

1. What does it mean for a group to be abelian?

Being abelian means that the group follows the commutative property, where the order in which elements are multiplied does not affect the result. In other words, the group operation is commutative.

2. How can you prove that a group is abelian?

A group can be proved to be abelian by showing that its operation is commutative. This can be done by demonstrating that for any two elements a and b in the group, a*b = b*a.

3. What are the properties of an abelian group?

Aside from the commutative property, an abelian group also has the identity element, inverse elements, and closure under the group operation. It also follows the associative property and has finite order.

4. Can a non-abelian group become abelian?

No, a non-abelian group cannot become abelian. The commutative property is inherent in the structure of an abelian group, and a non-abelian group cannot change its structure to become abelian.

5. Why is it important to determine if a group is abelian?

Determining if a group is abelian helps in understanding the structure and properties of the group. It also helps in solving problems and making calculations involving the group operations more efficient, as the commutative property makes the order of operations irrelevant.

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