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Thread: Number of Homomorphisms from Zn to Zm

  1. #1
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    Number of Homomorphisms from Zn to Zm

    Let $\displaystyle $Hom$\{\mathbb{Z}_n, \mathbb{Z}_m\}$ denote the group of homomorphisms between $\displaystyle \mathbb{Z}_n$ and $\displaystyle \mathbb{Z}_m$. Show that $\displaystyle |$Hom$\{\mathbb{Z}_n, \mathbb{Z}_m\}|=$gcd$(m,n)$.

    The only thing I have is that since we're dealing with a homomorphism that $\displaystyle |\phi(a)|$ divides $\displaystyle |a|$, but I don't see how that fact gets used to prove the above statement.
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  2. #2
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    Quote Originally Posted by mathematicalbagpiper View Post
    Let $\displaystyle $Hom$\{\mathbb{Z}_n, \mathbb{Z}_m\}$ denote the group of homomorphisms between $\displaystyle \mathbb{Z}_n$ and $\displaystyle \mathbb{Z}_m$. Show that $\displaystyle |$Hom$\{\mathbb{Z}_n, \mathbb{Z}_m\}|=$gcd$(m,n)$.

    The only thing I have is that since we're dealing with a homomorphism that $\displaystyle |\phi(a)|$ divides $\displaystyle |a|$, but I don't see how that fact gets used to prove the above statement.
    for an $\displaystyle f \in $Hom$\{\mathbb{Z}_n, \mathbb{Z}_m\}$ suppose $\displaystyle f(1+n\mathbb{Z})=k + m \mathbb{Z}.$ then $\displaystyle f(r+n\mathbb{Z})=kr + m\mathbb{Z}.$ for all $\displaystyle r.$ show that $\displaystyle f$ is well-defined if and only if $\displaystyle \frac{m}{d} \mid k,$ where $\displaystyle d=\gcd(m,n).$
    Last edited by NonCommAlg; Sep 9th 2010 at 04:40 PM.
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