# Math Help - Cross Cancellation implies commutativity?

1. ## Cross Cancellation implies commutativity?

Hello!

If we have a group G: if $a,b,c \in G$
and $ab=ca$ implies $b=c$

then prove G is Albelian.

To me, it seems trivial, that if we assume ab = ca and since b=c, $ab = ba$ $\forall a,b \in G$

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!

2. Originally Posted by matt.qmar
Hello!

If we have a group G: if $a,b,c \in G$
and $ab=ca$ implies $b=c$

then prove G is Albelian.

To me, it seems trivial, that if we assume ab = ca and since b=c, $ab = ba$ $\forall a,b \in G$

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!
You have not proven that, for all a,b in G, there exists c in G such that ab=ca.

(Whether or not we want to show this, or whether it's true or easy, I'm not sure. I'm not an expert on group theory, just happened to spot the error.)

3. 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!

In the assumption, we are saying G is a group with the property that,

IF ab=ca for some a,b,c in G

THEN b=c.

5. Originally Posted by matt.qmar
So I thought we could start by assuming ab = ca.
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:

Originally Posted by matt.qmar

In the assumption, we are saying G is a group with the property that,

IF ab=ca for some a,b,c in G

THEN b=c.
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.

6. Aha! I see how you mean.

Ok, I just thought...

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

7. Originally Posted by matt.qmar
Aha! I see how you mean.

Ok, I just thought...

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 $c=aba^{-1}$. 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.

8. That is quite clear to me now! Thank you!

9. So this is how it looks...

let e be the identity of G, let $a, b \in G$
$ab = ab(e) = ab(a^{-1}a)$
then by properties of G and commutativity
$a(b) = (aba^{-1})a$ implies $b = aba^{-1}$
right multiply by a, and commute
$ba = ab(a^{-1}a) = ab(e) = ab$
and so arbitrary a and b in G commute so it's Albelian.

I'm convinced. Thanks!!

10. Originally Posted by matt.qmar
So this is how it looks...

let e be the identity of G, let $a, b \in G$
$ab = ab(e) = ab(a^{-1}a)$
then by properties of G and commutativity
$a(b) = (aba^{-1})a$ implies $b = aba^{-1}$
right multiply by a, and commute
$ba = ab(a^{-1}a) = ab(e) = ab$
and so arbitrary a and b in G commute so it's Albelian.

I'm convinced. Thanks!!
You're welcome!

I think you meant associativity where you wrote commutativity -- commutativity is what we want to prove!

Also once you have that $b=aba^{-1}$ you can simply substitute to get $ab=ba$.

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 $a,b\in G$ (a and b not necessarily distinct), and let e be the identity element. Then

$ab=abe=aba^{-1}a=(aba^{-1})a$.

By the property of G, we have that $b=aba^{-1}$. Thus we substitute to get

ab=ba.

Thus G is abelian. QED.