Algebra Solutions: Laws of exponents in a group

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

Prove that $x^{a+b} = x^a x^b$ and $x^{ab} = (x^a)^b$ .
Prove that $(x^a)^{-1} = x^{-a}$ .
Prove part (1) when $a$ and $b$ are arbitrary integers.

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 $b$ .
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 $a \geq 1$ .
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 $x^{ab} = x^{(-a)(-b)} = (x^{-a})^{-b}$ $= (((x^a)^{-1})^{-1})^b$ $= (x^a)^b$ .

Algebra Solutions

Laws of exponents in a group

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