Exercise 3.1.20 ($ (\mathbb{Z}/24 \mathbb{Z})/ (12\mathbb{Z}/24\mathbb{Z}) \simeq \mathbb{Z}/12\mathbb{Z}.$)

Proof. Here $G=\mathbb{Z}∕24\mathbb{Z}$. The subgroup $H=⟨\overline{12}⟩=\{\bar{0},\overline{12}\}$ has order $2$. We write $\tilde{c}=\mathit{cH}$ the coset of $c\in G$ modulo $H$, and simplify the notation by writing $\widetilde{\stackrel{¯}{a}}$ as $ã$.

(a)

Let $\bar{a}$ be some element of $G$, where $a\in \mathbb{Z}$. The Euclidean division gives $q$ and $r$ such that $a=12q+r,0\le r<12$, so $\bar{a}=\overline{12}\bar{q}+\bar{r}$. Taking the classes modulo $H$, this gives, using $\widetilde{12}=\tilde{0}$, $$ã=\widetilde{12}\tilde{q}+\tilde{r}=\tilde{r},0\le r<12.$$

So

$$\tilde{G}=\{\tilde{0},\tilde{1},\dots ,\widetilde{11}\}.$$

Moreover, $|\tilde{G}|=|G|∕|H|=24∕2=12$, so all these elements are distinct.

(b)

For every integer $k$, $$k\tilde{1}=\tilde{0}⟺\tilde{k}=\tilde{0}⟺\bar{k}\in H=12\mathbb{Z}⟺12\mid k.$$

Therefore, $|\tilde{1}|=12$, and $|\tilde{k}|=|k\cdot \tilde{1}|=\frac{12}{12\wedge k}$. Explicitly,

$x$	$\tilde{0}$	$\tilde{1}$	$\tilde{2}$	$\tilde{3}$	$\tilde{4}$	$\tilde{5}$	$\tilde{6}$	$\tilde{7}$	$\tilde{8}$	$\tilde{9}$	$\widetilde{10}$	$\widetilde{11}$
$|x|$	$1$	$12$	$6$	$4$	$3$	$12$	$2$	12	$3$	$4$	$6$	$12$

(c)

We know that $\tilde{G}=⟨\tilde{1}⟩$ and $12\cdot \tilde{1}=\tilde{0}$. Since $Z_{12}=⟨x\mid x^{12}=1⟩$, there is a surjective homomorphism $\phi =Z_{12}\to \tilde{G}$ such that $\phi (x)=\tilde{1}$. Moreover $|Z_{12}|=|\tilde{G}|=12$, therefore $\phi :x^k\mapsto \tilde{k}$ is an isomorphism, so $$\tilde{G}\simeq Z_{12}\simeq \mathbb{Z}∕12\mathbb{Z}.$$

(Alternatively, we may verify that

$$\psi \{\begin{array}{ccc}\hfill \tilde{G}\hfill & \hfill \to \hfill & \mathbb{Z}∕12\mathbb{Z}\hfill \\ \hfill ã\hfill & \hfill \mapsto \hfill & [a{\text{]}}_{12}\hfill \end{array}$$

is well defined (if $\bar{a},\bar{b}\in G$, then $\bar{a}-\bar{b}\in H=⟨\overline{12}⟩\Rightarrow {[a]}_{12}={[b]}_{12}$), and that $\phi$ is an isomorphism.)

Note: Since $H=12\mathbb{Z}∕24\mathbb{Z}=\{\bar{0},\overline{12}\}$, this shows that

$$(\mathbb{Z}∕24\mathbb{Z})∕(12\mathbb{Z}∕24\mathbb{Z})\simeq \mathbb{Z}∕12\mathbb{Z}.$$

This is a particular case of the Third Isomorphism Theorem (see Section 3.3).