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Elementary Group Theory, p-group automorphism.

  1. Aug 11, 2011 #1
    I am self-studying elementary abstract algebra over the summer, with the book "Abstract Algebra: An Introduction" by Hungerford. It's my first exposure to mathematical proofs (well, short of an introduction in my intro to discrete mathematics), so sometimes I'm not really sure I do everything right in an exercise, and have nobody to ask or verify my work. This is one of those exercises. It's pretty early in the section, and I seemed to have a bit trouble proving it, so I want to ask here if my proof is correct/how I could have made it easier.

    1. The problem statement, all variables and given/known data
    If G is an abelian p-group and (n,p) = 1 prove that the map f: G -> G given by f(a) = na is an isomorphism.

    2. Relevant equations

    3. The attempt at a solution
    So what I need to prove in order for f to be an isomorphisms is that;
    1. f is a homomorphism
    2. f is injective
    3. f is surjective

    Everything below is in custom additive notation.

    1. This is trivial, as clearly
    f(a +b) = n(a + b) = a + b + ... + a + b = a + a + ... + a + b + .... + b = f(a) + f(b).
    Here I just used that in abelian groups, the operation defined is commutative.

    2. Injectivenes I prove by the fact that (n,p) = 1, and since every element a of G has an order that is a power of p, na != 0 if a != 0. Therefore, we see that the kernel of the homomorphism is just the identity element, and a previous result in the book states that this happens if and only if f is injective.

    3. Surjective
    Since for any element b that is in G, we know that the order of b is p^m for some integer m.
    Also, since (n,p) = 1, we know that (n, p^m) = 1. But that means that we have a solution in x,y to the equation (p^m)x + ny = 1. Multiplying by b I get b(p^m)x + bny = b. Now I use commutative law of the group and rewrite equation into:
    (bp^m)x + n(by) = b from which it follows that (since the first term becomes 0)
    n(by) = f(by) = b.

    It's the last part I'm a bit shaky about. Could someone tell me if this proof is correct, or in what way I should do it differently?
  2. jcsd
  3. Aug 11, 2011 #2
    This seems to be ok! :smile:


    I feel that you actually used associativity here, instead of commutativity.
  4. Aug 11, 2011 #3
    Hello, thanks for such a fast reply. It feels better to go on doing further exercises when I know I've done that one correctly.

    Oh, yes it's associativity indeed. I checked my post for errors twice and somehow missed that one :)
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