1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ $km = nz$ 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
  • 1
    $\begingroup$ 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
4
$\begingroup$

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 $\varphi(0) = 0$.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.