Exercise 1.1.6 (Examples of subgroups of $\mathbb{Q}$)

Proof. These sets $G$ are subsets of $\mathbb{Q}$ which contain $0$, so $\mathit{𝑡h𝑒𝑙𝑎𝑤𝑖𝑠𝑎𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑣𝑒},\mathit{𝑎𝑛𝑑𝑖𝑛𝑒𝑎𝑐h𝑐𝑎𝑠𝑒},-x\in G$ if $x\in G$. hence it is sufficient to verify that $G$ is closed under addition.

(a)

Let $G_1$ be the set of rational numbers (including $0=0∕1$) in lowest terms whose denominators are odd.

Let $x=a∕b$ and $y=c∕d$ be elements of $G_1$, where $g.c.d(a,b)=g.c.d(c,d)=1$ and where $b,d$ are odd.

$\frac{a}{b}+\frac{c}{d}=\frac{\mathit{𝑎𝑑}-\mathit{𝑏𝑐}}{\mathit{𝑏𝑑}}=\frac{e}{f}$, where $bg.c.d(e,f)$ is odd. Then $f$ divides $\mathit{𝑏𝑑}$, where $\mathit{𝑏𝑑}$ is odd, therefore $f$ is odd, so $x+y\in G_1$. This shows that $G_1$ is closed under addition, so $G_1$ is a group.

(b)

Let $G_2$ be the set of rational numbers (including $0=0∕1$) in lowest terms whose denominators are even.

Then $x=1∕6\in G_2$, $y=1∕10\in G_2$, but

$$\frac{1}{6}+\frac{1}{10}=\frac{4}{15},$$

where $4∕15$ is in lowest terms and $15$ is odd, so $x+y\notin G_2$. So $G_2$ is not a group.

(c)

Let $G_3$ be the set of rational numbers of absolute value $<1$.

Then $1∕2\in G_3$ and $2∕3\in G_3$, but $1∕2+2∕3=7∕6>1$, thus $G_3$ is not a group.

(d)

Let $G_4$ be the set of rational numbers of absolute value $\ge 1$ together with $0$.

Then $x=5∕3\in G_4$ and $y=-4∕3\in G_4$, but $x+y=1∕3\notin G_4$, thus $G_4$ is not a group.

(e)

Let $G_5$ be the set of rational numbers with denominators equal to $1$ or $2$.

Then $G_5$ is the set of half-integers:

$$G_5=\{n∕2\mid n\in \mathbb{Z}\}.$$

If $x=n∕2$ and $y=m∕2$ are any elements of $G_5$, then $x+y=(n+m)∕2\in G_5$, so $G_5$ is a group.

(f)

Let $G_6$ be the set of rational numbers with denominators equal to $1,2$ or $3$.

Then $x=1∕2\in G_6$ and $y=1∕3\in G_6$, but $x+y=5∕6\notin G_6$, so $G_6$ is not a group.