Hello!
If we have a group G: if
and implies
then prove G is Albelian.
To me, it seems trivial, that if we assume ab = ca and since b=c,
I have a feeling this proof is not valid, but I am not sure how? Does anyone else see an error in this logic, or have a suggestion for a method to prove this properly? Or is this legit?
Thank you!
Well I thought perhaps that given three arbitrary elements of G (a,b, and c)
to show the group is albelian, show ab = ba for two arbitrary elements.
So I thought we could start by assuming ab = ca. Then by the assumption of G, we know b =c, so we substitute b for c in the RHS, and get ab = ba.
I think I see the problem, though : when we say ab = ca, then a,b, and c are no longer arbitrary?
I think the way to do it is to start with any a and b,
then show ab must be = ba...
Sorry, hope this was clear! Thanks for the reply!
By doing this, you assume that such c exists, for arbitrary a and b. If there is no such element c in G for any particular a and b, then your proof falls apart.
Like I said, I'm not sure how to do the actual proof, just explaining how your reasoning has a gap in it.
Edit: We posted at the same time and I didn't see this until after submitting:
I'm still not sure we're on the same page, so consider this. IF we can prove that there always exists such a c, then we can write a full proof as such.
Let a, b in G. Then, by magical lemma, we know there exists c in G such that ab=ca. This implies that b=c, hence ab=ba, hence G is abelian. QED.
But if there are groups in which we are not guaranteed c exists (that is, we do not have our magical lemma to rely upon), then this proof does not work and we have to find another way.
Aha! I see how you mean.
Ok, I just thought...
start with ab = ab.
Then cabc = cabc (left and right multiply by some c in G)
then by the property of G, abc = cab
is this sufficent? since ab is in G (G is closed)
then call ab = d
so we have dc = cd, so G is commutative?
but now we have the condition d need be a product of elements, although each element in a group is a product of two other elements
Hmm I'm not sure... I got a little lost at the end.
But I think I just realized something obvious. I couldn't remember (!) whether we had inverses at our disposal when working in a group. But checking the axioms -- we do! So going back to the magical lemma proof in post #5, we can just let . I think we're done then right?
Edit: Looking at your latest proof (post #6) again, I believe it is valid, but has confusing lettering and presentation. If you did some re-lettering and added parentheses in a few places I think it would be easier to follow. Although I think it's cleaner to do it the "magical lemma" way, which is in fact very close to what you had in your post #1 anyway.
You're welcome!
I think you meant associativity where you wrote commutativity -- commutativity is what we want to prove!
Also once you have that you can simply substitute to get .
If you don't mind, an alternate way of stating the proof (I drop in and out of LaTeX just for ease of typing):
Let (a and b not necessarily distinct), and let e be the identity element. Then
.
By the property of G, we have that . Thus we substitute to get
ab=ba.
Thus G is abelian. QED.