# Show something's a homomorphism

1. Sep 28, 2006

### Nusc

LetH= {expim*theta; theta in R}. Show thatthe map G -> H, exp(i*theta) |-> exp(im*theta) is a group homomorphismi if and only if m is an integer.

(=>) Assume G - > is a group homo s.t. phi(exp(I*theta)) = (exp(I*m*theta))

Consider exp(ix), exp(iy) in G

Since G is a homo,

phi(exp(ix)exp(iy)) = phi(exp(ix))phi(exp(iy)))
phi(exp(i(x+y))) = phi(exp(ix))phi(exp(iy)))
exp(im(x+y) =exp(imx)exp(imy)

Isthe m on the left equal to the m on the right? I don't think so, what's wrong here?

How do I show the converse?

2. Sep 29, 2006

### matt grime

The way you've written it the m's must be the same. There is only one m kicking about the place.

You're attempting to show that raising z to z^m is a group homomorphism if and only if m is an integer (z in the complex circle |z|=1).

One way is obvious, and strangely it's the way you've not addressed: (wz)^m=w^mz^m 1^m=1 (if m is an integer, if it isn't then raising things to the fractional powers isn't necessarily well defined is it? i.e. do we pick 1 or -1 as the square root of 1? Hey, perhaps that idea will help with the other direction....)

So, for the direction you were writing. -1 is exp(1pi), -1 squared is 1, if m were 1/2, then (-1)^1/2 would'n't square to 1, would it?