**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.19**

Let $G$ be a group, $x \in G$, and $a,b \in \mathbb{Z}^+$.

**(1)** Prove that $x^{a+b} = x^a x^b$ and $x^{ab} = (x^a)^b$.

**(2)** Prove that $(x^a)^{-1} = x^{-a}$.

**(3)** Prove part $(1)$ when $a$ and $b$ are arbitrary integers.

Solution:

**(1)** First we show that $x^{a+b} = x^a x^b$ by induction on $b$. For the base case, note that $x^{a+1} = x^a x$ for all $a$. For the inductive step, suppose $x^{a+k} = x^a x^k$ for all $a$ and $1 \leq k \leq b$. Then $$x^{a+(b+1)} = x^{(a+b)+1} = x^{a+b} x = x^a x^b x = x^a x^{b+1}.$$ So by induction, $x^{a+b} = x^a x^b$ for all $a$ and $b$.

We now show that $x^{ab} = (x^a)^b$ by induction on $b$. For the base case, note that $x^{a \cdot 1} = x^a = (x^a)^1$ for all $a$. For the inductive step, suppose $x^{ak} = (x^a)^k$ for all $a$ and $1 \leq k \leq b$. Then $$x^{a(b+1)} = x^{ab + a} = x^{ab}x^a = (x^a)^b x^a = (x^a)^{b+1}.$$ So by induction, $x^{ab} = (x^a)^b$ for all $a$ and $b$.

**(2)** We prove this using induction. For the base case $a = 1$, we have $x^1 \cdot x^{-1} = 1$, so that $x^{-1} = (x^1)^{-1}$. For the inductive step, suppose $x^{-a} = (x^a)^{-1}$ for some $a \geq 1$. Then $$x^{a+1} x^{-(a+1)} = xx^ax^{-a}x^{-1} = xx^{-1} = 1,$$ using the definition of $x^a$ for negative exponents. Thus $x^{-(a+1)} = (x^{a+1})^{-1}$.

By induction, $x^{-a} = (x^a)^{-1}$ for all $a \geq 1$.

**(3)** Part (1) yields the case $a,b > 0$. Now for all integers $a$ and $b$, we have $$x^{a+0} = x^a = x^a x^0$$ and $$x^{0+b} = x^b = x^0 x^b.$$ Without loss of generality, two cases remain: $a,b < 0$ and ($a > 0$ and $b < 0$). In the first case, we have $$x^{a+b} = (x^{-b-a})^{-1} = (x^{-b} x^{-a})^{-1} = (x^{-a})^{-1} (x^{-b})^{-1} = x^a x^b.$$ In the second case, we have three subcases: $|b| < a$, $|b| = a$, and $|b| > a$. In the first case, we have $a+b > 0$. Then $$x^{a+b} x^{-b} x^{-a} = x^{a+b-b} (x^{a})^{-1} = x^a (x^a)^{-1} = 1,$$ so by the uniqueness of inverses, $$(x^{a+b})^{-1} = x^{-b} x^{-a} = (x^a x^b)^{-1}.$$ So $x^{a+b} = x^a x^b$. If $|b| = a$ we have $$x^{a+b} = x^{a-a} = 0 = x^a x^{-a} = x^a x^{b}.$$ If $|b| > a$, we have $$x^{a+b} = (x^{-b-a})^{-1} = (x^{-b} x^{-a})^{-1} = (x^{-a})^{-1} (x^{-b})^{-1} = x^a x^b.$$ Thus $x^{a+b} = x^a x^b$ for all integers $a$ and $b$.

Next, we show that $x^{ab} = (x^a)^b$ for all integers $a$ and $b$. If either of $a$ or $b$ is zero, then $x^{ab} = (x^a)^b$ holds trivially. If $a > 0$ and $b > 0$, we showed this in Part (1). If $a > 0$ and $b < 0$, then $ab \leq 0$. Now $$x^{ab} = (x^{-1})^{a(-b)} = ((x^{-1})^a)^{-b} = ((x^a)^{-1})^{-b} = (x^a)^b.$$ Similarly if $a < 0$ and $b > 0$. If $a,b < 0$, then $$x^{ab} = x^{(-a)(-b)} = (x^{-a})^{-b} = (((x^a)^{-1})^{-1})^b = (x^a)^b.$$