Why 7/6 and not 1/6? There is no spoon I used a more convoluted method If f is the homomorphism that sends elements of Z_12…
Why 7/6 and not 1/6?
There is no spoon
I used a more convoluted method
If f is the homomorphism that sends elements of Z_12 into products of powers of roots of unity, there are 12 different possible selections for what f(1) equals. f(1) = e^(pii/6) may be the most obvious. since 1 generates Z_12 f(1) is all you need to define to define f, but when the group is not cyclic you need to use a more general method to find the homomorphism.
first you would have to find the basis of the group by representing it as products of groups of prime power order, in this case represent Z_12 as the direct sum of Z_3 and Z_4.
0 = (0, 0)
1 = (1, 1)
2 = (2, 2)
3 = (0, 3)
4 = (1, 0)
….
11 = (2, 3)
then you know that Z_12 as ordered pairs of elements Z_3 and Z_4 can be generated as linear combinations of (1, 0) and (0, 1).
So in order to define a function f that sends elements of Z_12 into products of roots of unity, you only need to define f((1, 0)) to be ANY 3rd root of unity and f((0, 1)) to be ANY fourth root of unity.
In my case I arbitrarily defined f((1,0)) = e^(2ipi/3) and f((0, 1)) = e^(2ipi/4), which gives f((1, 1)) (the generator of the cyclic group) to be f((1, 0)) * f((0, 1)) = e^(7ipi/6)
Since there are 3 third roots of unity and 4 fourth roots of unity, there are 3*4 = 12 possible different functions and hence twelve different representations of Z_12 using the exponential function (when the group is not cyclic, the homomorphisms will not be isomorphisms).
“Typhoid & swans - it all comes from the same place.”