After classifying cyclic groups, we can describe homomorphisms from cyclic groups with remarkable efficiency. The focus keyword homomorphisms between cyclic groups means that once the image of a generator is known, the whole homomorphism is known. If $G=\langle a\rangle$, every element of $G$ is a power of $a$, so a homomorphism must satisfy $f(a^k)=f(a)^k$. The only restriction is that the chosen image of $a$ must respect the order of $a$. This lesson turns the abstract homomorphism condition into a practical listing method for maps between finite cyclic groups and from $\mathbb{Z}$ to cyclic groups. :::theorem[Homomorphisms from a Cyclic Group] Let $G=\langle a\rangle$ be a cyclic group and let $f:G\to H$ be a homomorphism. Then $f$ is completely determined by $f(a)$. If $o(a)=n<\infty$, then $$\begin{equation*} o(f(a))\mid n. \end{equation*}$$ ::: :::proof Given that $G=\langle a\rangle$ and $f:G\to H$ is a homomorphism. To prove that $f$ is determined by $f(a)$ and that $o(f(a))\mid n$ when $o(a)=n$. Since $G=\langle a\rangle$, every element of $G$ has the form $a^k$ for some $k\in\mathbb{Z}$. By the power property of homomorphisms, $$\begin{equation*} f(a^k)=f(a)^k. \end{equation*}$$ Thus the value of $f$ on every element is determined once $f(a)$ is known. If $o(a)=n$, then $a^n=e$. Applying $f$ gives $$\begin{align*} f(a^n)&=f(e),\\ f(a)^n&=e_H. \end{align*}$$ Therefore the order of $f(a)$ divides $n$. Hence $f$ is determined by $f(a)$, and $o(f(a))\mid n$. $\square$ ::: To list homomorphisms from a finite cyclic group, first choose a generator of the domain. Then find all elements in the codomain whose orders divide the order of that generator. Each such choice gives a homomorphism, and no other choice can work. In additive notation, a homomorphism from $\mathbb{Z}_n$ to $\mathbb{Z}_m$ is usually written by specifying $f([1])=[r]$. The condition becomes $n[r]=[0]$ in $\mathbb{Z}_m$. :::example[Homomorphisms from Z Six to Z Four] Find all homomorphisms $f:\mathbb{Z}_6\to\mathbb{Z}_4$. The group $\mathbb{Z}_6$ is generated by $[1]$, and $o([1])=6$. Therefore $f$ is determined by $f([1])$, whose order must divide $6$. In $\mathbb{Z}_4$, the element orders are $$\begin{equation*} o([0])=1,\quad o([1])=4,\quad o([2])=2,\quad o([3])=4. \end{equation*}$$ Only orders $1$ and $2$ divide $6$. Hence $$\begin{equation*} f([1])=[0]\quad \text{or}\quad f([1])=[2]. \end{equation*}$$ Thus the homomorphisms are $f([a])=[0]$ and $f([a])=[2a]$. ::: :::example[Homomorphisms from Z to Z] Find all homomorphisms $f:\mathbb{Z}\to\mathbb{Z}$. Since $\mathbb{Z}=\langle 1\rangle$, the homomorphism is determined by $f(1)$. Let $f(1)=m$. Then for every $n\in\mathbb{Z}$, $$\begin{equation*} f(n)=f(n\cdot 1)=nm. \end{equation*}$$ Thus all homomorphisms are $$\begin{equation*} f_m(n)=mn,\qquad m\in\mathbb{Z}. \end{equation*}$$ The onto homomorphisms occur only when $m=1$ or $m=-1$, because only then does every integer appear as a multiple of $m$. ::: :::solved-problem Find all homomorphisms $f:\mathbb{Z}_8\to\mathbb{Z}_{12}$. ::: :::solution Let $f:\mathbb{Z}_8\to\mathbb{Z}_{12}$ be a homomorphism. The group $\mathbb{Z}_8$ is generated by $[1]$, and $o([1])=8$. Therefore $f$ is determined by $f([1])$, and the order of $f([1])$ must divide $8$. In $\mathbb{Z}_{12}$, the elements whose orders divide $8$ are those of order $1$, $2$, or $4$. We compute: $$\begin{align*} o([0])&=1,\\ o([6])&=2,\\ o([3])&=4,\\ o([9])&=4. \end{align*}$$ Therefore $$\begin{equation*} f([1])\in\{[0],[3],[6],[9]\}. \end{equation*}$$ Hence all homomorphisms are $$\begin{equation*} f([a])=[0],\quad f([a])=[3a],\quad f([a])=[6a],\quad f([a])=[9a]. \end{equation*}$$ $\square$ ::: :::solved-problem Find all onto homomorphisms $f:\mathbb{Z}\to\mathbb{Z}_6$. ::: :::solution Let $f:\mathbb{Z}\to\mathbb{Z}_6$ be a homomorphism. Since $\mathbb{Z}=\langle 1\rangle$, the map is determined by $f(1)$. For $f$ to be onto, the element $f(1)$ must generate $\mathbb{Z}_6$. The generators of $\mathbb{Z}_6$ are the residue classes $[r]$ with $\gcd(r,6)=1$. Therefore the generators are $$\begin{equation*} [1]\quad \text{and}\quad [5]. \end{equation*}$$ Thus the onto homomorphisms are $$\begin{equation*} f(n)=[n]\quad \text{and}\quad f(n)=[5n]. \end{equation*}$$ $\square$ ::: :::theorem[Onto Maps Between Finite Cyclic Groups] Let $G$ and $H$ be cyclic groups of orders $n$ and $m$, respectively. If $m\mid n$, then there exists an epimorphism from $G$ onto $H$. ::: :::proof Given that $G=\langle a\rangle$, $H=\langle b\rangle$, $|G|=n$, $|H|=m$, and $m\mid n$. To prove that there exists an epimorphism from $G$ onto $H$. Define $f:G\to H$ by $$\begin{equation*} f(a^k)=b^k. \end{equation*}$$ This definition is well-defined because $a^n=e_G$ and $m\mid n$ imply $b^n=e_H$. For integers $r,s$, $$\begin{align*} f(a^ra^s)&=f(a^{r+s})\\ &=b^{r+s}\\ &=b^rb^s\\ &=f(a^r)f(a^s). \end{align*}$$ Thus $f$ is a homomorphism. Since $b=f(a)$ and $b$ generates $H$, the image of $f$ is all of $H$. Hence $f$ is an epimorphism. $\square$ ::: :::tip For homomorphisms between cyclic groups, do not try to assign images to every element independently. Choose the image of one generator, check its order, and then let the homomorphism condition determine all other values. ::: :::exercise 1. Find all homomorphisms from $\mathbb{Z}_5$ to $\mathbb{Z}_{10}$. 2. Find all homomorphisms from $\mathbb{Z}_{20}$ to $\mathbb{Z}_{10}$. 3. How many onto homomorphisms are there from $\mathbb{Z}$ to $\mathbb{Z}_8$? ::: :::answer 1. Since $o([1])=5$, $f([1])$ must have order dividing $5$. In $\mathbb{Z}_{10}$ the choices are $[0]$, $[2]$, $[4]$, $[6]$, and $[8]$, so there are five homomorphisms. 2. Every element of $\mathbb{Z}_{10}$ has order dividing $10$, and $10\mid 20$. Thus $f([1])$ may be any element of $\mathbb{Z}_{10}$, giving ten homomorphisms. 3. The onto homomorphisms correspond to generators of $\mathbb{Z}_8$, namely $[1],[3],[5],[7]$. There are four. ::: :::faq[Frequently Asked Questions] Q: Why does the image of a generator determine a homomorphism? A: Every element of a cyclic group is a power or multiple of the generator, and homomorphisms preserve powers. Q: Can any element be chosen as the image of a finite generator? A: No. Its order must divide the order of the generator. Q: When is a map from $\mathbb{Z}$ to a cyclic group onto? A: It is onto exactly when the image of $1$ is a generator of the codomain. ::: :::call-to-action[Continue to Power Maps] subtitle: Study maps of the form $a\mapsto a^n$ and learn when they become homomorphisms or isomorphisms. button: Study Power Maps | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/homomorphisms-of-groups/power-maps-on-groups button-ghost: Review Cyclic Classification | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/homomorphisms-of-groups/classification-of-cyclic-groups :::