## Laws of exponents in a group

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.

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

• Ray  On December 28, 2010 at 7:53 am

In the proof of part 2, I’m not sure we can use part 1 since -a is not a positive integer. I think x^a x^-a is equal to 1 just by definition in Dummit and Foote where x^-n is defined as (x^-1)^n.

• nbloomf  On December 28, 2010 at 8:34 am

You’re right. Thanks!

• Ray  On December 28, 2010 at 11:01 am

Oops in part 3 we forgot to prove x^ab = (x^a)^b for all integers.

• nbloomf  On December 28, 2010 at 1:12 pm

D’oh! Thanks for the heads up.

• Ray  On December 28, 2010 at 7:31 pm

For the last case in your proof of 3, is that when a,b are both negative? How did you get x^ab = (x^(-1))^(-a)b when b<0? Thanks

• nbloomf  On December 28, 2010 at 11:25 pm

My correction was not complete. Thanks for checking- I tried to take a shortcut that ended up being wrong. I think it works now.

• Ray  On December 28, 2010 at 9:37 pm

I as able to prove the case when a,b < 0 to get the other side x^ab. It was quite tricky with lots of parenthesis. Thanks.

• Ray  On December 29, 2010 at 2:53 am

Here’s what I have for a,b 0 and (-a)(-b)=ab. (x^a)^b = [x^-(-a)]^-(-b)
=[(x^-a)^-1]^(-1)(-b) <0 >>
= [{(x^-a)^-1}^-b]^-1 <0 >>
= [(x^-a)^(-1*-b)]^-1 <>
= [{(x^-a)^(-b)}^-1]^-1 <0>>
=(x^-a)^(-b) <>
= x^(-a*-b) <0>>
= x^ab

This problem was a lot tougher than I thought, its amazing how we take for granted the laws of exponents. Thanks.

• Ray  On December 29, 2010 at 2:59 am

Here’s what I have for a,b negative. So -a,-b positive and (-a)(-b)=ab.
(x^a)^b = [x^-(-a)]^-(-b)
=[(x^-a)^-1]^(-1)(-b) by part 2 since -a>0
= [{(x^-a)^-1}^-b]^-1 by part 2 again with -b>0
= [(x^-a)^(-1*-b)]^-1 by Defn of x^-n
= [{(x^-a)^(-b)}^-1]^-1 by part 2 again for -b>0
=(x^-a)^(-b) since (g^-1)^-1 =g
= x^(-a*-b) by part 1 for -a,-b>0
= x^ab

with the reasoning