Date: 09/28/2001 at 10:14:53 From: Clara Subject: Abstract Algebra/Abelian Groups Let G be a group with the following property: If a, b and c belong to G and ab = ca, then b = c. Prove that G is Abelian.
Date: 09/28/2001 at 14:37:23 From: Doctor Paul Subject: Re: Abstract Algebra/Abelian Groups Hi Clara. Thanks for writing to Dr. Math. This is a neat problem that wasn't immediately clear to me, but after thinking about it for a few minutes, I saw how to do it. In problems like this, we have to start with something we know to be true. Consider the following: Let x and y be elements of G. Then clearly xyx = xyx Now use the fact that the group operation is associative to rearrange things a bit: x(yx) = (xy)x Finally, apply the unique property with which the group G has been endowed to obtain: yx = xy We can do this since x,y in G implies xy and yx are also elements of G (the group must be closed under the given binary operation). Thus the group G is Abelian, which was to be shown. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/
Date: 10/06/2002 at 07:44:18 From: Jozef Subject: I don't think your answer is right! Hi, I don't think the above answer is right; it should be like this: if: ab = bc => a^(-1)*a*b = a^(-1)*c*a => b = a^(-1)*c*a Now we have to prove that group is abelian, but we know that b=c, so b = a^(-1)*b*a but this cannot be true if the group is not abelian, and then: b = a^(-1)*b*a = b*a^(-1)*a = b, only if it is abelian. Jozef
Date: 11/02/2002 at 01:21:32 From: Doctor Mike Subject: Re: I don't think your answer is right! Hello Jozef, First a few preliminaries. Your message line which starts with "if: ab = bc" should start with "if: ab = ca" instead. Also, sorry for the delay in getting you this answer. We want to know about any possible error in our archives, and carefully consider each case pointed out to us. Many budding mathematicians are confused about what works in a proof, and what's not okay. Most of us have refined our proof technique by a combination of "reading" and "writing." We read and try to understand good proofs, and practice doing our own and seek constructive criticism. I'd like to help you see clearly why Paul's proof does the job, and where your suggested approach gets off track. FIRST, let's expand, explain, and annotate his given proof. The basic structure of the proof is this: 1. Assume x and y are ANY elements of G. 2. A bunch of logical stuff is written.... 3. Therefore, xy = yx. Since we are not assuming anything special about x and y in step 1, the result in step 3 is true for any possible choice of x and y in group G. Let's do that, then. Assume x and y are in G. Both xy and yx are elements in G by the closure axiom (defining property) of groups. The next step is to use a fact that you do NOT have to assume. The associativity axiom says that combining 3 group elements is the same whichever method is used. In symbols, write L(MR)=(LM)R where the capital letters stand for Left, Middle, and Right. Here comes the clever part. What if you choose L and R to both be the same element? Choosing L=x and R=x and M=y, associativity says x(yx)=(xy)x. Dr. Paul uses the word "rearrange" to point out that "xyx" equals both of those. So why is this clever? Our important assumption that if ab=ca then b=c has not been used yet. Do you see why the fact x(yx)=(xy)x is in the form ab=ca? The important thing is having an equation where the element on the far left is the same as the far right element. For such a situation, the given fact about our group G guarantees that the middle two elements (near the = sign) are the same. To summarize, whenever ab=ca is true, then b=c is also true. Now we just have to use this. We know that x(yx)=(xy)x is true, so we also know that (yx)=(xy) is true. That means xy=yx. This is what was required for step 3, so we are done. (The archived proof is a more compact version.) SECOND, for the approach suggested in your mail. Please read carefully the exact wording of what we need to prove here. It never says that ab=ca is true. What it says is that "if" you ever are presented with a situation where ab=ca, then you are guaranteed that something else is also true, namely that b=c. Your subtle slip-up happens at the beginning, when you seem to take ab=ca as a starting point for logical derivations. Much of of what you write is true, but doesn't contribute to the desired result. FINALLY, I'd just like to confess to you and the world that many times I have initially thought a proof was faulty, only to realize my error after several hours (or days) of thought. Sometimes it also happens that published proofs ARE false. I have believed for a long time that the best way to learn a subject is to proof-read the textbook. Hope this has helped. Write back if you want to talk more about this proof, or proofs in general. Thanks for writing. - Doctor Mike, The Math Forum http://mathforum.org/dr.math/
Date: 11/02/2002 at 11:24:43 From: Jozef Subject: Thank you (I don't think your answer is right!) That 'trick' with associativity wasn't clear to me at first glance, but then I realized that it's very nice. Thanks a lot for your profound answer! Jozef
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum