Exercise 2.3.9 (Homomorphisms from $\mathbb{Z}/48\mathbb{Z}$ into $Z_{36}$)

Proof. Let $a\in \mathbb{Z}$. Suppose that there is a homomorphism ${\psi }_a:\mathbb{Z}∕48\mathbb{Z}\to Z_{36}$ such that ${\psi }_a(\bar{1})=x^a$. Since ${\psi }_a$ is a homomorphism, ${\psi }_a(\bar{k})=x^{\mathit{ak}}$ for all $k\in \mathbb{Z}$.

Indeed, ${\psi }_a(\bar{0})=1=x^0$, and if ${\psi }_a(\bar{k})=x^{\mathit{ak}}$ for some $k\ge 0$, then ${\psi }_a(\overline{k+1})={\psi }_a(\bar{k}+\bar{1})={\psi }_a(\bar{k}){\psi }_a(\bar{1})=x^{\mathit{ak}}{x}^a=x^{a(k+1)}$. The induction is done, which prove that for all integers $k\ge 0$, ${\psi }_a(\bar{k})=x^{\mathit{ak}}$.

Moreover, for $k\ge 0$, ${\psi }_a(-k)={({\psi }_a(k))}^{-1}={(x^{\mathit{ak}})}^{-1}=x^{a(-k)}$, so

For $k=48$, this gives

Since $|x|=36$, and $x^{48a}=1$, we obtain $36\mid 48a$, therefore $3\mid 4a$, where $3\wedge 4=1$, hence

Conversely, suppose that $3\mid a$. Then $a=3\alpha$ for some integer $\alpha$. Consider the map

• $\psi$ is well defined: If $\bar{k}=\bar{l}(k,l\in \mathbb{Z})$, then $l=k+48\lambda$, where $\lambda \in \mathbb{Z}$. Since $x^{36}=1$,

  $$x^{3\mathit{\alpha l}}=x^{3\alpha (k+48\lambda )}=x^{3\mathit{\alpha k}}{x}^{3\cdot 12\cdot 4\lambda }=x^{3\mathit{\alpha k}}{(x^{36})}^{4\lambda }=x^{3\mathit{\alpha k}},$$

  thus $x^{\mathit{ak}}=x^{3\mathit{\alpha k}}$ depends only of the class of $a$, so $\psi$ is well defined.

• $\psi$ is a homomorphism: for all $\bar{k},\bar{l}\in \mathbb{Z}∕48\mathbb{Z}$,

  $$\begin{array}{llll}\hfill \psi (\bar{k}+\bar{l})& =x^{a(k+l)}& \hfill & \\ \hfill & =x^{\mathit{ak}}{x}^{\mathit{al}}& \hfill & \\ \hfill & =\psi (\bar{k})\psi (\bar{l}).& \hfill & \end{array}$$
• Finally $\psi (\bar{1})=x^{a\cdot 1}=x^a$.

In conclusion, there is a homomorphism ${\psi }_a:\mathbb{Z}∕48\mathbb{Z}\to Z_{36}$ such that ${\psi }_a(\bar{1})=x^a$ if and only if $a$ is a multiple of $3$.

Note that ${\psi }_a$ is never surjective, otherwise $x\in {\psi }_a(\mathbb{Z}∕48\mathbb{Z})$, so $x={\psi }_a(\bar{k})=x^{3\mathit{\alpha k}}$ for some integer $k$ (where $a=3\alpha ,\alpha \in \mathbb{Z}$). This gives $x^{3\mathit{\alpha k}-1}=1$, where $|x|=3$, therefore $3\mid 3\mathit{\alpha k}-1$, so $3\mid 1$. This is false, so ${\psi }_a$ is not surjective. □