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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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    But I think it's safe to say that $\phi$ is a homomorphism as long as m is an integer.2012-10-31

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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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    @Fredrik_Meyer Thank you!2012-11-01