Dan,
Suppose is an isomorphism. Let x = (0,1). Then . So in
Since is an isomorphism, x=0, contradiction.
I need to show that is not isomorphic to .
Mind you, I'm in the section of my book dealing with cyclic groups. I was able to handle the other problems in this set by showing that one group is cyclic and the and the other isn't, etc. But isn't cyclic and I don't see how would be either. Any thoughts?
Thanks!
-Dan
IF the two were isomorphic so we had an isomorphism , then would necessarily be a cyclic subgroup of of order 2 under rational addition. Do the rational numbers possess such an additive subgroup (that is, is there ANY rational number q not equal to 0 with: q + q = 0)?
In other words, one group has a finite cyclic subgroup, the other does not. A similar proof holds for , for any positive integer n.
For starters f(k) isn't a homomorphism: For x, y in Z_4 we have
f(xy) = xy + 1
f(x)f(y) = (x + 1)(y + 1) = (xy + 1) + (x + y)
I can't seem to construct a bijective function g(k). The best I can do using the homomorphism property is g(k) = g^k(1), a cyclic group with generator g(1). But g(1) can only take the values 2, 3, or 4 and none of them work.
-Dan
The operation in Z_{4} is ADDITION.
However, it IS true f isn't a homomorphism:
f(x + y) = x + y + 1
f(x)f(y) = (x + 1)(y + 1) = xy + x + y + 1, and since 5 is prime, Z_{5} is an integral domain, so if x,y are non-zero, so is xy.
The two possible isomorphisms are:
f(k (mod 4)) = 2^{k} (mod 5) and
g(k (mod 4)) = 3^{k} (mod 5)
for example:
f(k (mod 4) + m (mod 4)) = f(k+m (mod 4)) = 2^{k+m} (mod 5) = (2^{k} (mod 5))(2^{m} (mod 5)) = f(k (mod 4))f(m (mod 4)).
The group (Z_{5}*, .) is also known as the group of units of Z_{5}, or U(5). In this group, 2 and 3 are primitive elements (generators). This is in direct analogy to the primitive 4-th roots of unity, as both 2 and 3 are square roots of -1 (which is 4 in Z_{5}):
(x - 2)(x - 3) (mod 5) = x^{2} - (2 + 3)x + 6 (mod 5) = x^{2} + 1 (mod 5).