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?
what is the necessary and sufficient condition needed?
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0We 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? – 2012-10-31
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0Do you assume that $\phi$ is a homomorphism? Do we have this fact for $\phi$? – 2012-10-31
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0No we don't assume that. – 2012-10-31
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0But I think it's safe to say that $\phi$ is a homomorphism as long as m is an integer. – 2012-10-31
1 Answers
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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0I've edited my answer slightly (it was wrong). Have you heard about the structure theorem of finitely-generated abelian groups? – 2012-10-31
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0I've never heard about it but I just looked it up and it looks very helpful! Thanks! – 2012-10-31
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1@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. – 2012-10-31
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0@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? – 2012-10-31
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0@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. – 2012-10-31
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0@Allison: I've expanded the answer somewhat with an example. Does it clarify? – 2012-10-31
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0@Fredrik_Meyer Thank you! – 2012-11-01