I am struggling with a problem from Artin's algebra. This is the solution I have so far.

Let $G,H$ be cyclic groups, generated by elements $x,y$. Determine the condition on the orders $m,n$ of $x$ and $y$ so that the map sending $x^i$ to $y^i$ is a group homomorphism.

Solution. Define the map $\varphi: G \to H$ with $x^i \mapsto y^i$. For this map to be well-defined, we require that $x^i = x^j$ imply that $y^i = y^j$, i.e., that $i \equiv j \ \text{(mod $m$)}$ implies that $i \equiv j \ \text{(mod $n$)}$. Stated differently, \begin{align*} \exists k \in \mathbb{Z}, \; i - j = km \implies \exists z \in \mathbb{Z}, \; i - j = nz. \end{align*} Hence, $km = nz$. But that would imply both that $m \mid n$ and $n \mid m$, meaning $n = m$, which is not true in general, of course. (In other words, I cannot figure out what to do with this conclusion, especially since these are two different sides of an implication.)

The homomorphism property would require $\varphi(xy) = \varphi(x)\varphi(y)$ for all $x,y \in G$. So $y^{i+m} = y^i y^m$, which is always true, so that doesn't seem to give us anything.

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    $\begingroup$ $km = nz$ does not imply that $m | n$ and $n | m$. Take eg $k=1$, $m = 6$, $n = 3$ and $z = 2$. $\endgroup$ – Olivier Roche Feb 17 at 8:44
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    $\begingroup$ Everything is correct until you come to $km=nz$. implies $n=m$. this is false. you made a mistake. $\endgroup$ – Baby desta Feb 17 at 8:47
  • $\begingroup$ It seems that it does imply that $n \mid m$, though, right? I suppose I am having trouble understanding why, even though that counterexample makes sense. (I'm sorry if this is obvious.) $\endgroup$ – John P. Feb 17 at 8:55
  • $\begingroup$ @JohnP. Indeed it implies $n | m$, see my answer ;) $\endgroup$ – Olivier Roche Feb 17 at 8:56
  • $\begingroup$ @JohnP. Yes it implies $m | n$. but not the other way around. the conclusion should be $m$ is a factor of $n$ $\endgroup$ – Baby desta Feb 17 at 8:58

A necessary condition for this map (let's call it $\varphi$) to be a morphism is that $n | m$, since we have $y^m = \varphi( x^m) = \varphi(0) = 0$.

It is also a sufficient condition since :

  • it ensures that the map is well defined.
  • it ensures that $\varphi(0) = 0$.

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