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Abelian Groups

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   

Date: 10/06/2002 at 07:44:18
From: Jozef
Subject: I don't think your answer is right!


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.


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 

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 

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   

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!

Associated Topics:
College Modern Algebra

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