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If G is a finite abelian group of order n and $\phi$ : G $\rightarrow $ G is defined by $\phi (a) = a^m \forall a \in $ G, what is the necessary and sufficient condition that guarantees that $\phi $ is an isomorphism of G onto itself?

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We don't really know it's an isomorphism, we're supposed to figure out what would turn it into an isomorhpism. Does that make sense? –  user39794 Oct 31 '12 at 18:22
    
Do you assume that $\phi$ is a homomorphism? Do we have this fact for $\phi$? –  Babak S. Oct 31 '12 at 18:25
    
No we don't assume that. –  user39794 Oct 31 '12 at 18:32
    
But I think it's safe to say that $\phi$ is a homomorphism as long as m is an integer. –  user39794 Oct 31 '12 at 18:43
    
+1 for nice question. :) –  Babak S. Oct 31 '12 at 19:00

1 Answer 1

up vote 1 down vote accepted

If $G$ is a finite abelian group, then $G$ is up to isomorphism on the form $\mathbb{Z}_{p_1^{n_1}} \times \mathbb{Z}_{{p_2}^{n_2}} \times \cdots \times \mathbb{Z}_{{p_n}^{n_n}}$, where $p_i$ are prime numbers (not necessarily distinct). Then your map takes $(1,1,\cdots,1)$ to $(m,m,\cdots,m)$.

For example, if $p_1^{n_1}|m$, then the map is zero on first factor, so in that case, the map is not an isomorphism. Can you see how this generalizes?

We can make this more concrete by an example. If $G=\mathbb{Z}_4$, then the map $\phi$ is just $1 \mapsto m \cdot 1$. If $m=2$, then $\phi$ has a non-trivial kernel: for $\phi(2)=2\cdot 2 = 4 = 0$. In fact, you can show that if $m$ and $4$ is not coprime (i.e. they have a common factor), then the map is not injective.

(also, nice fact: an injective map on finite sets is actually bijective. Can you prove this?)

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I've edited my answer slightly (it was wrong). Have you heard about the structure theorem of finitely-generated abelian groups? –  Fredrik Meyer Oct 31 '12 at 17:37
    
I've never heard about it but I just looked it up and it looks very helpful! Thanks! –  user39794 Oct 31 '12 at 17:40
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@AllisonCameron: I see this answer is nicely arranged. Consider the kernel of $\phi$. It is all elements $g\in G$ that $mg=0$ so, if you look for an injection, you should have $(m,|G|)=1$. I think, this is what Fredrik noted "For example, if.... is not isomorphism". And this finalizes your answer because as above, every injective between two finite structures is surjective as well. +1 for the answer. –  Babak S. Oct 31 '12 at 18:59
    
@BabakSorouh I still don't get it though, why do m and n need to be relatively prime for phi to be an injection? What happens if they aren't? –  user39794 Oct 31 '12 at 19:06
    
@AllisonCameron: Didn't you want $G\cong G$ under $\phi$? Indeed you wanted. Doesn't $\frac{G}{\ker\phi}\leq G$ when $\phi$ is an homomorphism?? If $(m,n)\neq 1$ so the kernel would be a proper subgroup of $G$ and so we wouldn't reach to our goal. Remember that if the order of an element $g\in G$ be relatively prime to $|G|$ then $g=0$ in an abelian group. –  Babak S. Oct 31 '12 at 19:14

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