I am trying to do this homework : Describe

$$\mbox{Hom}_{\mbox{Grp}} \left({\mathbb{Z}}/{n \mathbb{Z}}, {\mathbb{Z}}/{m \mathbb{Z}} \right).$$

My road(s) so far :

I note $\pi_n : \mathbb{Z} \to \mathbb{Z}/n \mathbb{Z}$ the canonical projection that sends $a\in\mathbb{Z}$ to $[a]_n$, its class on $\mathbb{Z}/n \mathbb{Z}$ and $\pi_m : \mathbb{Z} \to \mathbb{Z}/m \mathbb{Z}$ that sends $a\in\mathbb{Z}$ to $[a]_m$.

My first idea : Since $[1]_n$ spans $\mathbb{Z}/n \mathbb{Z}$, then for $\varphi \in \mbox{Hom}_{\mbox{Grp}} \left(\mathbb{Z}/n \mathbb{Z}\,,\, \mathbb{Z}/m \mathbb{Z} \right)$, $\varphi ([1]_n)$ must span $\mathbb{Z}/m \mathbb{Z}$ i.e. $m\varphi ([1]_n)=[0]_m$ which means that $\varphi([m]_n)=[0]_m$, as $\varphi$ is a group homomorphism. And then I am stuck because $[0]_n$ verify this condition.... and $m=0$ in $\mathbb{Z}/n \mathbb{Z}\iff m\in n\mathbb{Z}\iff m$ is a multiple of $n$. So $\varphi$ is a group homomorphism if and only if $m$ is a multiple of $n$ ? I don't think so...

My second idea : $[1]_n$ is order $n$ in $\mathbb{Z}/n \mathbb{Z}$, so $n[1]_n=[0]_n$. For $\varphi \in\mbox{Hom}_{\mbox{Grp}} \left(\mathbb{Z}/n \mathbb{Z}\,,\, \mathbb{Z}/m \mathbb{Z} \right)$, $\varphi([0]_n)=[0]_m$ because $\varphi$ is a group homomorphism. Then $\varphi(n[1]_n)=[0]_m$, which means that $\varphi([1]_n)$ is order $\varphi(n)$ in $\mathbb{Z}/m \mathbb{Z}$. And then, what next ? I can see where I'm going...

Thanks for your help

  • $\begingroup$ $\phi([1]_n$ doesn't need to span $\mathbb Z/m \mathbb Z$ it needs to span $\mathrm{im} \phi$. Also see. $\endgroup$ – JSchlather Jan 17 '13 at 19:56
  • $\begingroup$ @JacobSchlather Ok, my first idea is going to trash... Your link is talking about ring homomorphism, not group homomorphism $\endgroup$ – Alan Simonin Jan 17 '13 at 19:59
  • $\begingroup$ Think about the orders of the kernels. $\endgroup$ – Alexander Gruber Jan 17 '13 at 20:08
  • $\begingroup$ @AlexanderGruber the kernel is $n\mathbb{Z}$ for $n\in \mathbb{Z}$, no ? $\endgroup$ – Alan Simonin Jan 17 '13 at 22:50
  • $\begingroup$ See math.stackexchange.com/questions/273169 $\endgroup$ – Watson Jan 19 '17 at 20:54

It might be helpful to first show $\operatorname{hom}_\mathbb{Z}(\mathbb{Z}_{p^\ell},\mathbb{Z}_{p^k})\simeq\mathbb{Z}_{p^{\min\{\ell,k\}}}$.

Using your notation, denote elements of $\mathbb{Z}_{p^\ell}$ and $\mathbb{Z}_{p^k}$ by $[\cdot]_\ell$ and $[\cdot]_k$, respectively. Then define $$ f_{[\cdot]_k}\colon \mathbb{Z}_{p^\ell}\to\mathbb{Z}_{p^k}:[a]_\ell\mapsto a[x]_k. $$ Let $X=\{f_{[x]_k}:\operatorname{ord}([x]_k)\mid p^\ell\}$. I claim that $X=\operatorname{hom}_\mathbb{Z}(\mathbb{Z}_{p^\ell},\mathbb{Z}_{p^k})$. First, take $f_{[x]_k}\in X$. To see it is well defined, suppose $[a]_\ell=[b]_\ell$, so that $p^\ell\mid a-b$. Then $$ f_{[x]_k}([a]_\ell)-f_{[x]_k}([b]_\ell)=a[x]_k-b[x]_k=(a-b)[x]_k=0 $$ since $\operatorname{ord}([x]_k)\mid p^\ell$, and hence also divides $a-b$. Further, $f_{[x]_k}$ is also a homomorphism, since \begin{align*} f_{[x]_k}([a]_\ell+[b]_\ell) &= f_{[x]_k}([a+b]_\ell)\\ &= (a+b)[x]_k\\ &= a[x]_k+b[x]_k\\ &= f_{[x]_k}([a]_\ell)+f_{[x]_k}([b]_\ell). \end{align*} Conversely, Take $f\in\operatorname{hom}_\mathbb{Z}(\mathbb{Z}_{p^\ell},\mathbb{Z}_{p^k})$. For any $[a]_\ell\in\mathbb{Z}_{p^\ell}$, $$ f([a]_\ell)=f(a[1]_\ell)=af([1]_\ell) $$ and thus $f=f_{f[1]_\ell}$ and since $\operatorname{ord}(f[1]_\ell)\mid\operatorname{ord}[1]_\ell=p^\ell$, and so $f\in X$ and set equality follows.

I now claim $\vert X\vert=p^{\min\{\ell,k\}}$. This is equivalent to counting the number of elements of $\mathbb{Z}_{p^k}$ with order dividing $p^\ell$. If $k\leq\ell$, it follows that there are $p^k$ such elements by Lagrange's theorem. If $\ell\leq k$, then recall from group theory that $[x]_k=x[1]_k$ as order $p^k/(p^k,x)$. This divides $p^\ell$ if and only if $(p^k,x)/p^{k-\ell}$ is an integer. This occurs if and only if $p^{k-\ell}\mid x$, and there are $p^\ell$ such element in $\mathbb{Z}_{p^k}$ since we can take $x=p^{k-\ell}\cdot c$ for $c=1,\dots,p^\ell$. In either case, $\vert X\vert=p^{\min\{k,\ell\}}$.

I now claim $\operatorname{ord}([x]_k)=\operatorname{ord}(f_{[x]_k})$, viewing $X$ as an additive group. Observe $$ 0=\operatorname{ord}(f_{[x]_k})f_{[x]_k}([a]_\ell)=\operatorname{ord}(f_{[x]_k})a[x]_k $$ for all $a$. So for $a=1$, we conclude $\operatorname{ord}([x]_k)\leq\operatorname{ord}(f_{[x]_k})$. Also, $$ \operatorname{ord}([x]_k)f_{[x]_k}([a]_\ell)=\operatorname{ord}([x]_k)\cdot a[x]_k=0 $$ so $\operatorname{ord}(f_{[x]_k})\leq\operatorname{ord}([x]_k)$. Now $[p^{k-\min\{\ell,k\}}]_k$ has order $\min\{\ell,k\}$, and thus the corresponding element of $X$ does, and so $X\simeq \mathbb{Z}_{p^{\min\{\ell,k\}}}$.

Also, observe that $\operatorname{hom}_\mathbb{Z}(\mathbb{Z}_{p^i},\mathbb{Z}_{q^j})\simeq\{0\}$ if $p\neq q$, since the image of any element under a homomorphism must then divide $p^i$ and $q^j$, and thus has order $1$, from which it follows that the only homomorphism between the two groups is the trivial homomorphism.

Now you can adapt this to arbitrary $m$ and $n$.

So write $n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ and $m=p_1^{\beta_1}\cdots p_k^{\beta_k}$, where the exponents may be $0$. Then recalling how the $\operatorname{hom}$ functor splits over products and direct sums, which can be found in Lang's Algebra, Chapter III, Section $3$, we have \begin{align*} \operatorname{hom}_\mathbb{Z}(\mathbb{Z}_n,\mathbb{Z}_m)&\simeq \operatorname{hom}_\mathbb{Z}\left(\prod_{i=1}^k\mathbb{Z}_{p_i^{\alpha_i}},\prod_{j=1}^k\mathbb{Z}_{p_j^{\beta_j}}\right)\\ &\simeq \prod_{i,j=1}^k\operatorname{hom}_\mathbb{Z}\left(\mathbb{Z}_{p_i^{\alpha_i}},\mathbb{Z}_{p_j^{\beta_j}}\right)\\ &\simeq \prod_{\ell=1}^k\operatorname{hom}_\mathbb{Z}\left(\mathbb{Z}_{p_\ell^{\alpha_\ell}},\mathbb{Z}_{p_\ell^{\beta_\ell}}\right)\\ &\simeq \prod_{\ell=1}^k\mathbb{Z}_{p_\ell^{\min\{\alpha_\ell,\beta_\ell\}}}\\ &\simeq \mathbb{Z}_{p_1^{\min\{\alpha_1,\beta_1\}}\cdots p_k^{\min\{\alpha_k,\beta_k\}}}\\ &=\mathbb{Z}_{(n,m)}. \end{align*}

(I read this proof a few years back while studying group theory, and TeXed it up. I can't recall the original source however, apologies!)

  • $\begingroup$ Is there a reason why the spoiler markup isn't working? $\endgroup$ – Ben West Jan 17 '13 at 20:53
  • $\begingroup$ Thanks for your answer and for typing it in a Latex format ! I'll look at it $\endgroup$ – Alan Simonin Jan 17 '13 at 21:45
  • $\begingroup$ Maybe should I prove the isomorphism by giving a bijective group homomorphism ? $\endgroup$ – Alan Simonin Jan 19 '13 at 0:06
  • 1
    $\begingroup$ @AlanSimonin The last paragraph already shows that $\operatorname{hom}_{\mathrm{Grp}}(\mathbb{Z}/n\mathbb{Z},\mathbb{Z}/m\mathbb{Z})$ is isomorphic to $\mathbb{Z}/(m,n)\mathbb{Z}$, where $(m,n)=\gcd(m,n)$. $\endgroup$ – Ben West Jan 19 '13 at 6:23
  • $\begingroup$ Yes indeed. As I can see, you construct a group isomorphism with the $f_{[\dot]_k}$. I think there can be a simplier proof than this one... I search in this direction. Thanks again ! $\endgroup$ – Alan Simonin Jan 19 '13 at 14:26

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