12 Oct 07
Originally posted by TheMaster37How 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?
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.
I don't like cheap counter-examples. I always miss them and do copious amounts of hard work instead...
14 Oct 07
2 edits
Originally posted by geniusI suspect you mean that there is no element a, not equal to 0, so that a*b = 0 for some b?
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 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.