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Barre
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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.
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.
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?
Homework Statement
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.
Homework Equations
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?
