How many subgroups $K \le \mathbb{Z}_n$ are there with $K =\ker(\phi)$ for some homomorphism $\phi\colon\mathbb{Z}_n \rightarrow \mathbb{Z}_m$?

Stuck and need a hint. Have so far that there are $\gcd(n,m)$ such homomorphisms. Also that all such maps are of the form $\phi_a(k)= ak$ for $a \in \mathbb{Z}_m$ satisfying $na=0 \pmod m$.

Specifically $a = 0, \dfrac{m}{d}, \dfrac{2m}{d},... (d-1)\dfrac{m}{d}$, where $d = gcd(n,m)$.

Thus we can denote $\ker(\phi_i)$ by set of all $k$ where $ki\dfrac{m}{d}=0 \pmod m$ for $i \in \mathbb{Z}_d$. Further I showed that if $i,j \in U_d$ then $\ker(\phi_i)=\ker(\phi_j)$. However I am stuck with figuring out how many kernels there are for remaining $d-|U_d|$ homomorphisms(aside from trivial map).

  • $\begingroup$ Hint: $\mathbb{Z}_n$ is a cyclic group, so $\phi$ is uniquely determined by the value $\phi(1)$. $\endgroup$ – Secret Math Jun 27 '13 at 5:04
  • $\begingroup$ If $K$ is a subgroup of $\mathbb{Z}_n$, then there is a homomorphism $\mathbb{Z}_n\to\mathbb{Z}_n/K$. What can you say about the codomain? $\endgroup$ – egreg Jun 27 '13 at 6:33
  • $\begingroup$ That it is the left cosets of K in $Z_n$. And we know |aK|=|K|. So $|Img(\phi)|=|K|$. Isn't this just giving me the number of elements in the kernal? Or is this giving the number of distinct subgroups which are kernels of this type of homomorphism? $\endgroup$ – Tim Jun 27 '13 at 7:24
  • $\begingroup$ Is $m$ fixed? By the First Isomorphism Theorem, you know that the kernel of homomorphisms are exactly the normal subgroups. But since $\mathbb{Z}_n$ is abelian all subgroups $K$ are normal, so that I can find a homomorphism with kernel equal to $K$... or maybe I am missing something? $\endgroup$ – Ryan Sullivant Jun 29 '13 at 23:46
  • $\begingroup$ @RyanSullivant yes, for this question $m$ is fixed. $\endgroup$ – leo Jun 30 '13 at 0:23

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