Exercise 2.1.1 (Examples of subgroups)

Proof.

(a)

Put $G_a=\{a+\mathit{ai}\mid a\in ℝ\}$.

• $G_a\subset ℝ$ and $0=0+0i\in G_a$, thus $G_a\ne \varnothing$.
• If $z=a+\mathit{ai}\in G_a$ and $t=b+\mathit{bi}\in G_a$, then $z-t=(a-b)+i(a-b)=c+\mathit{ic}$, where $c=a-b\in ℝ$, so $z-z^{\prime }\in G_a$.

$G_a$ is a subgroup of $ℝ$.

(b)

Put $G_b=𝕌=\{z\in ℂ\mid |z|=1\}$.

• $|1|=1$ so $1\in 𝕌$, thus $𝕌\ne \varnothing$.
• If $z,t\in 𝕌$, then $|z|=|t|=1$. Therefore $|z∕t|=|z|∕|t|=1$, so $z{t}^{-1}\in 𝕌$.

$G_b=𝕌$ is a subgroup of $ℂ$.

(c)

(I don’t know what is the denominator of a rational number: I assume that this is the denominator of the reduced fraction representing this rational.)

Let $n$ be a positive integer. Put

$$G_c=\{x\in \mathbb{Q}\mid \exists <!--\; FUNCTION\; APPLICATION\; -->p\in \mathbb{Z},\exists <!--\; FUNCTION\; APPLICATION\; -->q\in \mathbb{Z},q>0,x=p∕q,p\wedge q=1\text{ and }q\mid n\}.$$

• $G_c\subset \mathbb{Q}$ and $1=1∕1\in G_c$, thus $G_c\ne \varnothing$.

• Let $x,y$ be elements of $G_c$, where $x=a∕b,(a,b)\in {\mathbb{Z}}^2,b>0,a\wedge b=1,b\mid n$ and $y=c∕d,(c,d)\in {\mathbb{Z}}^2,d>0,c\wedge d=1,d\mid n$.

  Since $b\mid n$ there is some integer $k>0$ such that $n=\mathit{kb}$, and similarly there is some integer $l>0$ such that $n=\mathit{ld}$. Then

  $$x-y=\frac{a}{b}-\frac{c}{d}=\frac{\mathit{ka}}{\mathit{kb}}-\frac{\mathit{lc}}{\mathit{ld}}=\frac{\mathit{ka}-\mathit{lc}}{n}.$$

  Let $e∕f$ be the reduced fraction representing $x-y$, so that $x-y=e∕f,(e,f)\in {\mathbb{Z}}^2,f>0,e\wedge f=1$. Then

  $$x-y=\frac{e}{f}=\frac{\mathit{ka}-\mathit{lc}}{n}.$$

  This implies that $\mathit{en}=f(\mathit{ka}-\mathit{lc})$, thus $f\mid \mathit{en}$, where $f\wedge e=1$, therefore $f\mid n$. This shows that $x-y\in G_c$.

$G_c$ is a subgroup of $\mathbb{Q}$.

(d)

Here $n$ is a fixed positive integer and $$G_d=\{x\in \mathbb{Q}\mid \exists <!--\; FUNCTION\; APPLICATION\; -->p\in \mathbb{Z},\exists <!--\; FUNCTION\; APPLICATION\; -->q\in \mathbb{Z},q>0,x=p∕q,p\wedge q=1\text{ and }q\wedge n=1\}.$$

• $G_d\subset \mathbb{Q}$ and $1=1∕1\in G_d$, thus $G_d\ne \varnothing$.

• Let $x,y$ be elements of $G_d$, where $x=a∕b,(a,b)\in {\mathbb{Z}}^2,b>0,a\wedge b=1,b\wedge n=1$ and $y=c∕d,(c,d)\in {\mathbb{Z}}^2,d>0,c\wedge d=1,d\wedge n=1$. Then

  $$x-y=\frac{a}{b}-\frac{c}{d}=\frac{\mathit{ad}-\mathit{bc}}{\mathit{bd}}.$$

  Since $b\wedge n=1$ and $d\wedge n=1$, then $\mathit{bd}\wedge n=1$. Let $e∕f$ be the reduced fraction representing $x-y$, so that $x-y=e∕f,(e,f)\in {\mathbb{Z}}^2,f>0,e\wedge f=1$. Then

  $$x-y=\frac{e}{f}=\frac{\mathit{ad}-\mathit{bc}}{\mathit{bd}}.$$

  This implies that $\mathit{ebd}=f(\mathit{ad}-\mathit{bc})$, so $f\mid e(\mathit{bd})$, where $f\wedge e=1$, therefore $f\mid \mathit{bd}$. Since $\mathit{bd}\wedge n=1$, a fortiori $f\wedge n=1$. This shows that $x-y\in G_d$.

$G_d$ is a subgroup of $\mathbb{Q}$.

(e)

Put $G_e=\{x\in {ℝ}^{\text{×}}\mid x^2\in \mathbb{Q}\}$.

• $G_e\subset {ℝ}^{\text{×}}$ and $1^2=1\in \mathbb{Q}$, thus $1\in G_e$ and $G_e\ne \varnothing$.
• If $x,y\in G_e$, put $a=x^2$ and $b=y^2\ne 0$, then $a,b\in \mathbb{Q}$, thus ${(x{y}^{-1})}^2=x^2∕y^2=a∕b\in \mathbb{Q}$, therefore $x{y}^{-1}\in G_e$.

$G_d$ is a subgroup of ${ℝ}^{\text{×}}$.