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Thread: Isomorphism

  1. #1
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    Question Isomorphism

    If G is a finite abelian group of order n and φ: G -> G’ is defined by φ(a)= a^m for all a element of G find the necessary and sufficient condition that φ be an isomorphism of G itself
    Last edited by Godisgood; Oct 11th 2009 at 06:26 AM.
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  2. #2
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    Quote Originally Posted by Godisgood View Post
    If G is a finite abelian group of order n and φ: G -> G is defined by φ(a)= am for all a element of G find the necessary and sufficient condition that φ be an isomorphism of itself
    Start by rewriting the question! You are asserting that "φ be an isomorphism of itself" which makes no sense. Grammatically you want "φ be an isomorphism of G to itelf", also called an "automorphism".

    Also you have defined φ by "φ(a)= am" without saying what "m" is. A fixed member of G?
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  3. #3
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    I think you meant that $\displaystyle \phi$ acts on an element by applying the group operation m times. Now, in order to say that this forms an automorphism, you have to show that

    $\displaystyle (ab)^m = \phi(ab) = \phi(a)\phi(b) = a^m b^m$.

    This is true because the group is Abelian. Next, for it to be an automorphism it has to be onto. Since the group is finite, if you can show that $\displaystyle \phi$ is one-to-one, it would imply that it is onto.

    For it to be one-to-one, it is necessary that $\displaystyle \forall a,b \in G$ if $\displaystyle \phi(a) = \phi(b)$, then it must be true that $\displaystyle a=b$. Thus, $\displaystyle a^m = b^m$ implies that $\displaystyle a=b$. That would definitely be true if $\displaystyle (m,n)=1$, but that might be overkill. There might be a weaker condition that works.
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  4. #4
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    It occurs to me that $\displaystyle (m,n)=1$ isn't enough. In the group $\displaystyle U(9)$, $\displaystyle 4^2 = 16 = 7 (mod 9)$ and at the same time $\displaystyle 5^2 = 25 = 7 (mod 9)$, thus $\displaystyle \phi(a) = a^2$ is not one-to-one even though $\displaystyle (2,9) = 1$. But the rest of my argument holds. I just can't think of a sufficient condition. Sorry

    A condition that will absolutely work is if the order of the group is a prime number. then every element is a generator, which in turn means that the $\displaystyle \phi(a) = a^m$ is a generator as long as $\displaystyle m < n$. Can anyone think of a weaker condition?
    Last edited by eeyore; Oct 12th 2009 at 12:09 PM. Reason: Added another possible condition.
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