Closed, finite set under *.

Originally posted by TheMaster37
Alas, you still haven't seen my counterexample.

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

It is not closed. You don't have a product that results in B.

Originally posted by adam warlock
It is not closed. You don't have a product that results in B.

That doesn't matter. All that matters is that the result is contained in the set.

Originally posted by Palynka
That doesn't matter. All that matters is that the result is contained in the set.

You're right! And he's right too.

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 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 eleme ...[text shortened]... tity; there is no element X so that B*X = X*B = B.

Above is even a commutative closed group.

How about a set, as before, but with the conditionb that there is no element such that 0.a=a for all a in the set S?

I don't like cheap counter-examples. I always miss them and do copious amounts of hard work instead...

Originally posted by genius
How about a set, as before, but with the conditionb that there is no element such that 0.a=a for all a in the set S?

I don't like cheap counter-examples. I always miss them and do copious amounts of hard work instead...

I suspect you mean that there is no element a, not equal to 0, so that a*b = 0 for some b?

I still think there is a counterexample. Commutativity is a big thing to lose.

I'm trying to think of a group that works like this:

a*a = a
b*b = b
a*b = a
b*a = b

Then there would be no element 1 that satisfies
1*a = a*1 = a
AND
1*b = b*1 = b

EDIT:
Got it

1 1 = a
0 0

1 0 = b
0 0

The set { a , b } is closed under multiplication, has no 0, is not commutative and has no 1.

Originally posted by TheMaster37

EDIT:
Got it

1 1 = a
0 0

1 0 = b
0 0

The set { a , b } is closed under multiplication, has no 0, is not commutative and has no 1.

Nice work.