Closed, finite set under *.

Originally posted by adam warlock
And since you are so advanced here you go {1,a,1/a} with a different from 0,1, and -1.

That's not closed in general - e.g. a^2

Originally posted by mtthw
That's not closed in general - e.g. a^2

Yeah... 😳 That's not closed at all...

Originally posted by mtthw
Is it obvious you can do this? I could believe it, but I'm not sure it's that straightforward.

Just seen this post. But I didn't get it. What do you mean?

Originally posted by adam warlock
Just seen this post. But I didn't get it. What do you mean?

The additional bit you added to your proof so that you didn't need to use commutativity didn't seem obviously true to me. But I'm probably missing something simple.

Originally posted by mtthw
The additional bit you added to your proof so that you didn't need to use commutativity didn't seem obviously true to me. But I'm probably missing something simple.

It's not true in general.

There is no guarantee that

a[1] * ...* a[k-1] * a[k+1] * ... * a[n] * a[k] is equal to a[1] *...* a[n]

You NEED commutativity for that. Instead of reals, think of matrices. Those aren't commutative, but there are plenty of closed groups without a unity element.

Originally posted by TheMaster37
It's not true in general.

There is no guarantee that

a[1] * ...* a[k-1] * a[k+1] * ... * a[n] * a[k] is equal to a[1] *...* a[n]

You NEED commutativity for that. Instead of reals, think of matrices. Those aren't commutative, but there are plenty of closed groups without a unity element.

It is true in here. All I'm using is the fact that S is closed to * So I multplied all of the elements of the set and the product has to be an element of the set.

For instance if the set was {a,b,c} a*b*c=x where x belongs to S. But c*b*a=y where y still belongs to S. Now x and y may be different or equal depending on if * is commutative or not BUT the fact that the remaining factors equalls 1 when multiplied still holds.

So a[1]*...*a[k]*...*a[n]=a[k] and a[1]*...*a[k-1]*a[k+1]*...*a[n]*a[k]=a[k] are both valid. Now we can't say a priori if both a[k] are equal but the equality still holds and that's all we need to prove that the remaining factors multiply to get the identity. And we can use a[k] for both expressions cause k is just a dummy index.

I hope I made it clear enough. If not just tel me so.

Originally posted by adam warlock
It is true in here. All I'm using is the fact that S is closed to * So I multplied all of the elements of the set and the product has to be an element of the set.

For instance if the set was {a,b,c} a*b*c=x where x belongs to S. But c*b*a=y where y still belongs to S. Now x and y may be different or equal depending on if * is commutative or not BUT t sions cause k is just a dummy index.

I hope I made it clear enough. If not just tel me so.

The problem I have is that it might be a unit for this single element a[k].

For an element to be called 1, you need a * 1 = 1 * a = a for ALL elements a in the group.

Make two 2x2 matrices;

0 0 = A
0 0

0 1 = B
0 0


The set {A ,B} is closed under multiplication;

A*A = A
A*B = A
B*A = A
B*B = A

But neither of the two elements is the identity; there is no element X so that B*X = X*B = B.

Above is even a commutative closed group.

Originally posted by TheMaster37
The problem I have is that it might be a unit for this single element a[k].

For an element to be called 1, you need a * 1 = 1 * a = a for ALL elements a in the group.

Make two matrices;

A = 0 0 B = 0 1
0 0 0 0

The set {A ,B} is closed under multiplication;

A*A = A
A*B = A
B*A = A
B*B = A

But neither of the two ...[text shortened]... tity; there is no element X so that B*X = X*B = B.

Above is even a commutative closed group.

Once again k is a dummy index and we made no special hypothesis about the nature of a[k] so the equality is valid for all a[k]. That is to say that for all a[k] (k ranging from 1 to n) we prove that that is an identity element.

The question now is about the unicity of the identity. If * is commutative then the identity is unique if * isn't commutative then the identity needs not to be unique.

Originally posted by adam warlock
Once again k is a dummy index and we made no special hypothesis about the nature of a[k] so the equality is valid for all a[k]. That is to say that for all a[k] (k ranging from 1 to n) we prove that that is an identity element.

The question now is about the unicity of the identity. If * is commutative then the identity is unique if * isn't commutative then the identity needs not to be unique.

Nice dodge, I gave a counterexample, and yet you maintain that your proof is valid.

You prove that for every element in the group there is another that, upon multiplying, acts as a unit. That is not the same as proving the identity is in the group.

Originally posted by TheMaster37
You prove that for every element in the group there is another that, upon multiplying, acts as a unit. That is not the same as proving the identity is in the group.

Well, if the set is closed to * then a[1]*...*a[k-1]*a[k+1]*...*a[n] =A is a set element. And by A*a[k]=a[k] it equals the identity.

I think you haven't read all posts with enough attention.

Originally posted by adam warlock
So a[1]*...*a[k]*...*a[n]=a[k] and a[1]*...*a[k-1]*a[k+1]*...*a[n]*a[k]=a[k] are both valid. Now we can't say a priori if both a[k] are equal but the equality still holds and that's all we need to prove that the remaining factors multiply to get the identity. And we can use a[k] for both expressions cause k is just a dummy index.

I hope I made it clear enough. If not just tel me so.

I'm with you most of the way. But you do seem to be assuming that there is some sequence of the a[j]s which, when multiplied together, equals the last member of that sequence. I know it doesn't matter which sequence that is - once you've got one the rest of the proof follows. But the existence of one such sequence isn't obvious to me.

Originally posted by mtthw
I'm with you most of the way. But you do seem to be assuming that there is some sequence of the a[j]s which, when multiplied together, equals the last member of that sequence. I know it doesn't matter which sequence that is - once you've got one the rest of the proof follows. But the existence of one such sequence isn't obvious to me.

It is like that because the set S is closed to *. So if you multiply all members of the set the result must be some member of the set.

I put that member of the set at last place of the sequence because that way we can achive our proof without having to resort to commutativity.

Originally posted by adam warlock
It is like that because the set S is closed to *. So if you multiply all members of the set the result must be some member of the set.

I put that member of the set at last place of the sequence because that way we can achieve our proof without having to resort to commutativity.

Yes, it's got to be a member of the set. But as soon as you move the member to the last place of the sequence you are rearranging the sequence, and potentially changing the value of the product to a different member of the set. So you then have to shift a different value to the end...changing the result again.

Originally posted by mtthw
Yes, it's got to be a member of the set. But as soon as you move the member to the last place of the sequence you are rearranging the sequence, and potentially changing the value of the product to a different member of the set. So you then have to shift a different value to the end...changing the result again.

Yes. I see it now. Without assuming commutativity the proof seems to be flawed. I'll try to think it again and see if I can come to something. If not at least we have the commuting proof.

Originally posted by adam warlock
Yes. I see it now. Without assuming commutativity the proof seems to be flawed. I'll try to think it again and see if I can come to something. If not at least we have the commuting proof.

Alas, you still haven't seen my counterexample.

The set I gave is finite, closed and commutative. Yet no identity.