Question

prove or disprove
if an operation is commutative on a two element set, then it is associative on the same set.

Answer 1

i think I found a counter example

Answer 2

Here's what I'm thinking:

Let \(*:\{a,b\}\to\{a,b\}\) and \(a*b=b*a\).

To show associativity, you have to show \(a*(b*c)=(a*b)*c\) for \(a,b,c\in\{a,b\}\).

First case: Let \(c=a\). Then you have
\[a*(b*c)=a*(b*a)\\
(a*b)*c=(a*b)*a=a*(a*b)=a*(b*a)~~~~~\text{due to commutativity}\]
Second case: Let \(c=b\), then
\[a*(b*b)=\cdots\\
(a*b)*b=\cdots~~~~~\text{(whatever these may be)}\]
Does your example say that the last two (when \(c=b\)) aren't the same?

Answer 3

my counter example shows that there is an operation on a 2 element group that is commutative but not associative

Answer 4

I was going your rout but then there are so many cases...and its not true anyway:)

Answer 5

not group .... 2 element set

Answer 6

Well, there are really only two cases, as I outlined above. All you have to check is whether or not the operation you came up with shows that \(a*(b*c)\not=(a*b)*c\). In particular, when \(c=b\).

According to your Cayley table, you have
\[a*(b*b)=a*a=b\\
(a*b)*b=a*b=a\]
Not equal, so not associative.

Answer 7

right, thats exactly what I meant by a counter example:P