|Oct13-05, 06:18 PM||#1|
I want to find Aut(Z) so I listed the generators of Z which are 1, and -1. Then from the hint in the book it says once I know where the generators are mapped then I can get where every other element gets mapped. So I said the only choices for the automorphisms are defined by a(1) = 1 or a(1) = -1.
How do I know that these are the only automorphisms? If I could say that an automorphism maps a generator to a generator that would help, but I dont really know that and cant prove it right now. Any thoughts?
|Oct13-05, 07:09 PM||#2|
Automorphisms are homomorphisms, so if you know where an automorphism g sends a generator, you know where it sends everything (when the group has a single generator), since g(m) = g(m x 1) = m x g(1)*. Suppose g sends 1 to something other than 1 or -1. Well something must be sent to 1, since g is surjective. Suppose g(m) = 1. Then m x g(1) = 1, g(1) = 1/m, but 1/m is not even in Z unless m = 1 or m = -1. Check that the automorphisms you've defined are indeed automorphisms. In general, the mapping that sends x to its inverse is not an automorphism, but is in this case. See if you can figure out, in general, what conditions on the group must hold in order for such a mapping to be an automorphism.
* Actually, this looks a little confusing. Although when we speak of the group Z, we adopt additive notation (that is, we denote the group operation on x and y as x+y, whereas in a general group we would denote it as simply xy, i.e. in multiplicative notation). If we agree to use multiplicative notation when speaking of Z, then adding 1 to itself m times would be denoted as 1m, and adding x to y would be denoted xy. If we also agree to write elements of Z in bold, then what I wrote would look like this:
g(m) = g(1m) = g(1)m
In additive notation, it would look like:
g(m) = g(m x 1) or g(m1) = m x g(1) or mg(1)
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