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.

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Comments

  • 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

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