Exercise 2.3.8 (When does $\overline{1} \mapsto x^a$ extend to an isomorphism from $\mathbb{Z}/ 48 \mathbb{Z}$ onto $Z_{48}$?)

Proof. Suppose that there is some isomorphism ${\psi }_a$ 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 proves 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

Since ${\psi }_a$ is surjective, $x\in {\psi }_a(\mathbb{Z}∕48\mathbb{Z})$, so there is some integer $k$ such that $x={\psi }_a(\bar{k})=x^{\mathit{ak}}$. Therefore $x^{\mathit{ak}-1}=1$, where $|x|=48$, thus $48\mid \mathit{ak}-1$, so $\mathit{ak}-1=48l$, where $k,l$ are integers. The (Bézout’s) relation $\mathit{ak}-48l=1$ between $a$ and $48$ shows that

Conversely, suppose that $48\wedge a=1$, and consider the map

• $\psi$ is well defined: If $\bar{k}=\overline{k^{\prime }}$, where $k,k^{\prime }\in \mathbb{Z}$, then $k\equiv k^{\prime }(\mathit{mod}48)$, so $k^{\prime }=k+48\lambda$ for some $\lambda \in \mathbb{Z}$. Therefore

  $$x^{a{k}^{\prime }}=x^{\mathit{ak}}{(x^{48})}^{\mathit{a\lambda }}=x^{\mathit{ak}}.$$

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

  $$\begin{array}{llll}\hfill \psi (\bar{k}+\bar{l})& =x^{a(k+l)}& \hfill & \\ \hfill & =x^{\mathit{ak}}{x}^{\mathit{al}}& \hfill & \\ \hfill & =\psi (\bar{k})\psi (\bar{l}).& \hfill & \end{array}$$
• $\psi$ is injective: If $\bar{k}\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )$, then $x^{\mathit{ak}}=1$. Since $48\wedge a=1$, $|x^a|=\frac{48}{48\wedge a}=48$, and ${(x^a)}^k=1$, hence $48\mid k$, so $\bar{k}=\bar{0}.$ This shows that $\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )=\{\bar{0}\}$.

• $\psi$ is surjective: Let $y\in {\mathbb{Z}}_{48}=⟨x⟩$. Then $y=x^k$ for some integer $k$. Since $a\wedge 48=1$, there are integers $u,v$ such that $\mathit{au}+48v=1$, thus $a(\mathit{ku})+48(\mathit{kv})=k$, that is $\mathit{ar}+48s=k$, where $r=\mathit{ku},s=\mathit{kv}$ are integers. Then

  $$y=x^k=x^{\mathit{ar}+48s}=x^{\mathit{ar}}{(x^{48})}^s=x^{\mathit{ar}}=\psi (\bar{r})(\bar{r}\in \mathbb{Z}∕48\mathbb{Z}).$$

  (Alternatively, we may use Proposition 6 (2), or the argument of cardinality).

• Finally, $\psi (\bar{1})=x^a$.

Therefore $\psi :\mathbb{Z}∕48\mathbb{Z}\to Z_{48}$ is an isomorphism such that $\psi (\bar{1})=x^a$.

(Alternatively, if we know the first isomorphism Theorem of section 3.3, we factor the surjective homomorphism $\phi :\mathbb{Z}\to Z_{48}$ defined by $\phi (k)=x^{\mathit{ak}}$, which satisfies $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=48\mathbb{Z}$, to obtain the isomorphism $\psi :\mathbb{Z}∕48\mathbb{Z}\to Z_{48}.$)

In conclusion, the integers $a$ such that there is some isomorphism ${\psi }_a:\mathbb{Z}∕48\mathbb{Z}\to Z_{48}$ such that ${\psi }_a(\bar{1})=x^a$ are the integers $a$ relatively prime to $48$. □