Let be a group, , and .

- Prove that and .
- Prove that .
- Prove part (1) when and are arbitrary integers.

- First we show that by induction on . For the base case, note that for all . For the inductive step, suppose for all and . Then . So by induction, for all and .
We now show that by induction on . For the base case, note that for all . For the inductive step, suppose for all and . Then . So by induction, for all and .

- We prove this using induction. For the base case , we have , so that . For the inductive step, suppose for some . Then , using the definition of for negative exponents. Thus . By induction, for all .
- Part (1) yields the case . Now for all integers and , we have and . Without loss of generality, two cases remain: and and . In the first case, we have . In the second case, we have three subcases: , , and . In the first case, we have . Then , so by the uniqueness of inverses, . So . If we have . If , we have . Thus for all integers and .
Next, we show that for all integers and . If either of or is zero, then holds trivially. If and , we showed this in part 1. If and , then . Now . Similarly if and . If , then .

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## Comments

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.

You’re right. Thanks!

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

D’oh! Thanks for the heads up.

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

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

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.

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.

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