Simple Abstract Algebra Problem

[m0bius]

Homework Statement

Let G be a nonempty finite set closed under an associative operation such that both the left and right cancellation laws hold. Show that G under this operation is a group.

Homework Equations

My book defines the left and right cancellation laws as :

"For any a,b in G, the equations ax = b and you = b have unique solutions."

The Attempt at a Solution

I took 2 semesters of algebra using this book and now that I have some free time I've decided to relearn it. I think I'm on the right track but I'm stuck.

Anyway, to show G is a group we need to show that the inverse and identity axioms hold, since we are already given closure and associativity. Identity needs to be shown first since the inverse axiom uses the element e.

Identity
Let a,b be in G with a = b. Then we know by the left/right cancellation laws there exist two elements x,y in G such that ax = a and you = a, and these solutions are unique. This is where I'm stuck. If we can show x = y, then by definition x = y = e and we are done. I am having trouble doing this but I seem very close, so I feel like showing x = y is obvious but I'm just being stupid.. Is this even the right path?

Another option would be to show that in a set G as described in the problem, ax = a implies x = e. This is obviously true if G were a group, but we don't know that it is yet.


Inverse
I can't go on without assuming the identity axiom holds, so assume it does. Let a,b be in G where b = e and a is an arbitrary element in G. Then by the left/right cancellation laws, ax = e and you = e for some x,y in G. I seem to be stuck in the same way here too. If x = y then by definition x = y = a^-1 and we are done, but I can't seem to find a way to show this.

If it's even possible to give hints without giving the answer, any help would be appreciated. Thanks!

**On a side note, I tried using tex tags on my equations and for things like "a,b in G". Some of them worked but I'd get very odd errors everywhere else, like [*tex]a,b \in G[*/tex] (minus the *'s) coming out as "ya = b", and vice versa. Also every time you preview your post it slaps another template onto the bottom of it, anyway to stop that?

 

[Dick]

For the first one, multiply ax=a on the right by a. So you get axa=aa. Now left cancel an a.

 

[m0bius]

For the first one, multiply ax=a on the right by a. So you get axa=aa. Now left cancel an a.

Thank you, I haven't done this in a while...but that's really no excuse for something like this. I'm actually still confused about the hint you gave though, but it helped me get the inverse part. Isn't left/right "canceling" the same as multiplying by an inverse, which we don't know exists yet?
Meaning, couldn't we just left cancel ax = a to get x = e? (ie ax = a >>> a^-1ax = a^-1a >>> ex = e >>> x = e)

I guess I'm confused about what type of manipulation is allowed.

**the tex tags are still not working for me so ">>>" means "implies".

I think this is right for inverse:

We have ax = e and you = e.

ax = e >>> y(ax) = y(e) >>> (ya)x = ye >>> ex = ye >>> x = y. So by definition x = y = a^-1.

Thanks again,
m0bius.

 

[Dick]

Thank you, I haven't done this in a while...but that's really no excuse for something like this. I'm actually still confused about the hint you gave though, but it helped me get the inverse part. Isn't left/right "canceling" the same as multiplying by an inverse, which we don't know exists yet?
Meaning, couldn't we just left cancel ax = a to get x = e? (ie ax = a >>> a^-1ax = a^-1a >>> ex = e >>> x = e)

I guess I'm confused about what type of manipulation is allowed.

**the tex tags are still not working for me so ">>>" means "implies".

I think this is right for inverse:

We have ax = e and you = e.

ax = e >>> y(ax) = y(e) >>> (ya)x = ye >>> ex = ye >>> x = y. So by definition x = y = a^-1.

Thanks again,
m0bius.

It's not the same as assuming an inverse. You have a(xa)=aa and a(a)=aa so by uniqueness you have xa=a. That's why the uniqueness is called 'cancellation'. Then since you have xa=a and ya=a you can conclude x=y. That's a right cancellation by uniqueness. Now you've got that every element has an identity. But you haven't yet shown all elements have the same identity, have you?

 

[m0bius]

It's not the same as assuming an inverse. You have a(xa)=aa and a(a)=aa so by uniqueness you have xa=a. That's why the uniqueness is called 'cancellation'. Then since you have xa=a and ya=a you can conclude x=y. Now you've got that every element has an identity. But you haven't yet shown all elements have the same identity, have you?

Ahhh I understand the cancellation part now, I never understood that fully I guess. So now we have :

For each element a in G, there exists an element x such that ax = xa = a.

Now I'm a little confused here, please don't think I'm trolling or something, I just want to be 100% sure. I feel I'm making this more difficult than it needs to be, but is that statement above equivalent to :

There exists an element x in G such that for all a in G, ax = xa = a?

This second statement is the definition of the identity axiom as it is written in my book, and I feel like the two might be different. If they are the same, would we really need to show that x is unique? The axiom doesn't require that you show it is unique, even though it is.

 

[Dick]

Ahhh I understand the cancellation part now, I never understood that fully I guess. So now we have :

For each element a in G, there exists an element x such that ax = xa = a.

Now I'm a little confused here, please don't think I'm trolling or something, I just want to be 100% sure. I feel I'm making this more difficult than it needs to be, but is that statement above equivalent to :

There exists an element x in G such that for all a in G, ax = xa = a?

This second statement is the definition of the identity axiom as it is written in my book, and I feel like the two might be different. If they are the same, would we really need to show that x is unique? The axiom doesn't require that you show it is unique, even though it is.

I don't think you are trolling or anything. This stuff is tricky at first. And even later it can be tricky. But the two statements are different. One says every element of G has an identity and the second says for all elements of G the identity is the same. Axioms don't require you to show it's unique. But you can't treat this as an axiom. You have to SHOW there is unique identity. You really prove this a lot like you proved every element has an identity. I could feed you another hint, but why don't you just mess around and try it first now that you get cancellation.

 

[m0bius]

I don't think you are trolling or anything. This stuff is tricky at first. And even later it can be tricky. But the two statements are different. One says every element of G has an identity and the second says for all elements of G the identity is the same. Axioms don't require you to show it's unique. But you can't treat this as an axiom. You have to SHOW there is unique identity. You really prove this a lot like you proved every element has an identity. I could feed you another hint, but why don't you just mess around and try it first now that you get cancellation.

I'm not sure why I'm having difficulty with this, but I am. Right now I'm trying to show:

For any a,b in G, there exists identity elements x,y in G such that ax=xa=a and by=yb=b. Show x=y.

Is this the right path? We've already shown that such elements x and y exist, now we just need to show that they are the same for any two elements, and therefore the same for all elements.

 

[Dick]

I'm not sure why I'm having difficulty with this, but I am. Right now I'm trying to show:

For any a,b in G, there exists identity elements x,y in G such that ax=xa=a and by=yb=b. Show x=y.

Is this the right path? We've already shown that such elements x and y exist, now we just need to show that they are the same for any two elements, and therefore the same for all elements.

It's the cancellation trick again. Take ax=a. Multiply by b, so axb=ab. Try and take it from there.

 

[m0bius]

It's the cancellation trick again. Take ax=a. Multiply by b, so axb=ab. Try and take it from there.

Ok, I was multiplying ax = a by y and was getting nowhere. So,

ax = a >>> axb = ab >>> axb = ayb >>> xb = yb >>> x = y.

Hopefully that's right, I think it is but I've been looking at this problem too long and it's been confusing me since the start.

 

[Dick]

Ok, I was multiplying ax = a by y and was getting nowhere. So,

ax = a >>> axb = ab >>> axb = ayb >>> xb = yb >>> x = y.

Hopefully that's right, I think it is but I've been looking at this problem too long and it's been confusing me since the start.

Sure. That works. I was doing axb=ab -> xb=b. So x is the identity for b as well. But there's nothing wrong with what you did. Now that you have a unique identity e, the inverse part should be the EASY part. Look back the steps you've already done. It's just more of the same.

 

[m0bius]

I tried the inverse part earlier in the thread. First, using the cancellation laws, we let a = b. Then:

We have ax = e and you = e.

ax = e >>> y(ax) = y(e) >>> (ya)x = ye >>> ex = ye >>> x = y. So by definition x = y = a^-1.

It seems correct, and it's for any arbitrary element a in G so I think this is all we need to show.

 

[Dick]

I tried the inverse part earlier in the thread. First, using the cancellation laws, we let a = b. Then:
It seems correct, and it's for any arbitrary element a in G so I think this is all we need to show.

I was waiting until you knew you had a well defined element 'e'. Now that you know that, sure, it is that easy. 'e' is the hard part compared with that.