Exercise 1.1.19 ($x^{a+b} = x^a x^b$ and $(x^a)^b = x^{ab}$)

Proof. Let $x\in G$. By the inductive definition of $x^n$ (see p. 20), for all $n\in \mathbb{N}$,

(a)

Let $a\in \mathbb{N}$ be any fixed positive or zero integer. We proceed by induction on $b$.

• First, by (1),

  $$x^{a+0}=x^a=x^a\cdot x^0.$$

• Assume that for some $b\in \mathbb{N}$, $x^{a+b}=x^a{x}^b$. Then, using (2),

  $$x^{a+(b+1)}=x^{a+b}\cdot x=x^a{x}^{b}x=x^a{x}^{b+1}.$$

• The induction is done, which proves that for all $b\in \mathbb{N}$, $x^{a+b}=x^a{x}^b$. Since $a$ is an arbitrary integer $(a\in \mathbb{N})$, then

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->a\in \mathbb{N},\forall <!--\; FUNCTION\; APPLICATION\; -->b\in \mathbb{N},x^{a+b}=x^a{x}^b.$$

Similarly, let $a\in \mathbb{N}$ be any fixed positive or zero integer. We proceed by induction on $b$.

• By (1),

  $${(x^a)}^0=1=x^0=x^{a\cdot 0}.$$

• Assume ${(x^a)}^b=x^{\mathit{𝑎𝑏}}$ for some $b\in \mathbb{N}$. Then

  $$\begin{array}{lllllll}\hfill {(x^a)}^{b+1}& ={(x^a)}^b\cdot x^a& \hfill (\text{by (2)}& & \hfill & & \hfill \\ \hfill & =x^{\mathit{𝑎𝑏}}{x}^a& \hfill \text{(by the induction hypothesis)}& & \hfill & & \hfill \\ \hfill & =x^{\mathit{𝑎𝑏}+a}& \hfill \text{by the first part}& & \hfill & & \hfill \\ \hfill & =x^{a(b+1)}.& \hfill & & \hfill & \end{array}$$

• The induction is done, so

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->a\in \mathbb{N},\forall <!--\; FUNCTION\; APPLICATION\; -->b\in \mathbb{N},{(x^a)}^b=x^{\mathit{𝑎𝑏}}.$$

(b)

By (3), for any $a\in \mathbb{N}$, $$x^{-a}={(x^{-1})}^a.$$

We must prove ${(x^a)}^{-1}={(x^{-1})}^a$, or equivalently $x^a{(x^{-1})}^a=1$. By part (a), $x^a\cdot x=x^{a+1}=x^{1+a}=x^1{x}^a=x\cdot x^a$, so $x$ commutes with $x^a$.

• First $x^0{(x^{-1})}^0=1$.

• Assume $x^a{(x^{-1})}^a$ for some $a\in \mathbb{N}$ . Then ${(x^a)}^{-1}={(x^{-1})}^a$, thus

  $$\begin{array}{llllll}\hfill x^{a+1}{(x^{-1})}^{a+1})& =x^{a}x{(x^{-1})}^a{x}^{-1})& \hfill & & \hfill & \\ \hfill & =x^{a}x{(x^a)}^{-1}{x}^{-1}& \hfill & & \hfill & \\ \hfill & =x{x}^a{(x^a)}^{-1}{x}^{-1}& \hfill (\text{since }x{x}^a=x^{a}x)& & \hfill & & \hfill \\ \hfill & =x{x}^{-1}& \hfill & & \hfill & \\ \hfill & =1.& \hfill & & \hfill & \end{array}$$

• The induction is done which proves that for all $a\in \mathbb{N}$, $x^a{(x^{-1})}^a=1$, so

  $$\begin{array}{lll}\hfill x^{-a}={(x^{-1})}^a={(x^a)}^{-1}(a\in \mathbb{N}).& & \hfill \text{(4)}\end{array}$$

Note: It is important for part (c) to prove that (4) remains true if $a<0(a\in \mathbb{Z})$. Put $\alpha =-a$, so that $\alpha >0$. By equalities (4), where $a$ is replaced by $\alpha >0$,

$$x^{-\alpha }={(x^{-1})}^{\alpha }={(x^{\alpha })}^{-1}.$$

Therefore their inverses are equal:

$${\left(x^{-\alpha }\right)}^{-1}={\left({(x^{-1})}^{\alpha }\right)}^{-1}=x^{\alpha }.$$

By (4), where $x$ is replaced by $x^{-1}$ and $a$ by $\alpha >0$,

$${\left(x^{-1}\right)}^{-\alpha }=x^{\alpha }={\left({(x^{-1})}^{\alpha }\right)}^{-1}.$$

This shows that

$$x^{\alpha }={(x^{-1})}^{-\alpha }={(x^{-\alpha })}^{-1},$$

where $\alpha =-a$, so for any integer $a<0$

$$x^{-a}={(x^{-1})}^a={(x^a)}^{-1}.$$

Therefore (4) is true for any $a\in \mathbb{Z}$:

$$\begin{array}{lll}\hfill x^{-a}={(x^{-1})}^a={(x^a)}^{-1}(a\in \mathbb{Z}).& & \hfill \text{(5)}\end{array}$$

(c)

Now let $a,b$ be any integers.

• if $a\ge 0$ and $b\ge 0$, then by part (a),

  $$x^{a+b}=x^a{x}^b.$$

• if $a\ge 0$ and $b<0$, then $-b>0$, thus

  • If $a+b\ge 0$, then

    $$x^a=x^{a+b+(-b)}=x^{a+b}{x}^{-b},$$

    and since $x^{-b}={(x^b)}^{-1}$, then

    $$x^{a+b}=x^a{x}^b.$$

  • If $a+b<0$, then $-a-b>0$, and by (5) and part (a);

    $$x^a{(x^{a+b})}^{-1}=x^a{x}^{-a-b}=x^{-b}={(x^b)}^{-1},$$

    thus

    $$x^{a+b}=x^a{x}^b.$$

• If $a<0$ and $b\ge 0$, we prove by exchanging the roles of $a$ and $b$ that $x^{a+b}=x^a{x}^b$.

• If $a<0$ and $b<0$, then by (5)

  $$x^{a+b}={(x^{-1})}^{-a-b}={(x^{-1})}^{-a}{(x^{-1})}^{-b}=x^a{x}^b.$$

In every case,

$$x^{a+b}=x^a{x}^b(a,b\in \mathbb{Z}).$$

Similarly,

• If $a\ge 0$ and $b\ge 0$, then by part (a),

  $${(x^a)}^b=x^{\mathit{𝑎𝑏}}.$$

• If $a\ge 0$ and $b<0$, then using (5)

  $${(x^a)}^b={\left({(x^a)}^{-b}\right)}^{-1}={(x^{-\mathit{𝑎𝑏}})}^{-1}=x^{\mathit{𝑎𝑏}}.$$

• If $a<0$ and $b\ge 0$, then using (5)

  $${(x^a)}^b={\left({(x^{-1})}^{-a}\right)}^b={(x^{-1})}^{-\mathit{𝑎𝑏}}=x^{\mathit{𝑎𝑏}}.$$

• If $a<0$ and $b<0$, then

  $${(x^a)}^b={\left({\left({(x^{-1})}^{-a}\right)}^{-b}\right)}^{-1}={\left({(x^{-1})}^{(-a)(-b)}\right)}^{-1}={(x^{-\mathit{𝑎𝑏}})}^{-1}=x^{\mathit{𝑎𝑏}}.$$

In every case

$${(x^a)}^b=x^{\mathit{𝑎𝑏}}(a,b\in \mathbb{Z}).$$