# Exercise 2.4.11 in Artin

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.

• $km = nz$ does not imply that $m | n$ and $n | m$. Take eg $k=1$, $m = 6$, $n = 3$ and $z = 2$. – Olivier Roche Feb 17 at 8:44
• Everything is correct until you come to $km=nz$. implies $n=m$. this is false. you made a mistake. – Baby desta Feb 17 at 8:47
• 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.) – John P. Feb 17 at 8:55
• @JohnP. Indeed it implies $n | m$, see my answer ;) – Olivier Roche Feb 17 at 8:56
• @JohnP. Yes it implies $m | n$. but not the other way around. the conclusion should be $m$ is a factor of $n$ – 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 ensures that $$\varphi(0) = 0$$.