# proved that a group is abelian

by Benzoate
Tags: abelian, proved
 P: 569 1. The problem statement, all variables and given/known data Prove that a group G is Abelian if and only if (ab)^-1 =a^-1*b^-1 for all a and b in G. 2. Relevant equations No equations for this problem 3. 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
 Sci Advisor HW Helper P: 9,398 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?
P: 569
 Quote by matt grime 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

P: 299

## proved that a group is abelian

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 $$\left( ab \right)^{-1} = b^{-1}a^{-1}$$ (make sure you understand why) that's two one-liners.
HW Helper
P: 9,398
 Quote by Benzoate 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.
P: 299
I think I managed to understand it:
 Quote by Benzoate 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).
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,706 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.
P: 569
 Quote by Timo 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
 HW Helper P: 4,125 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
HW Helper
P: 9,398
 Quote by Benzoate 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.

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