Theorem 3.12 Parts (a), (c), & (d) in Gilbert –
Let G be a group.
(Part (a)) Show that for any x in G, xn x-n = e for any integer n.
Proof:
We will need the following related claim: x xn = xn x for any integer n.
Since x x1 = x x = x1 x, the claim holds if n = 1. Now assume it holds for
some k . Thenx xk+1 = x (xk x) = (x xk) x = xk+1 x, so it holds for k + 1
and therefore, the principle induction, it holds for all n > 0.
It’s clear that the claim holds for n = 0, since x x0 = x e = x = e x = x0 x.
Now recall that if n < 0, then xn is defined to be (x*)-n for x* the inverse of x.
This means x-1 is defined as (x*)-(-1) = x* which is consistent with our usage
to date of that symbol, i.e. x-1 for the inverse of x. Thus, the claim holds for
n = -1, since x x-1 = e = x-1 x.
Suppose n < 0. Then –n > 0 and per the above we have:
x xn = x (x*)-n = x (x*)-n (x* x) = x ((x*)-n x*) x =
x (x* (x*)-n) x = ((x x*)(x*)-n) x = (x*)-n x = xn x
Therefore, x xn = xn x holds for all integers n & any x in G.
Finally, as regards Part (a) –
For x in G, we have x0 x-0, so the claim holds if n = 0.
We also have that x1 x-1 = x x* = e, so it holds if n = 1.
Suppose now that the claim holds for some k > 0. Then
xk+1 x-(k+1) = (xk x) (x*)k+1 = (xk x) ((x*)k x*) = (xk x) (x* (x*)k) =
(xk (x x*)) (x*)k = (xk e) (x*)k = xk x-k = e.
So, by the principle of induction, it holds for all n ≥ 0.
If k < 0, then –k > 0 and, per what was just shown, we have
x-k x-(-k) = e, or x-k xk = e, which implies that xk x-k = e (why?).
Part (c) For any x in G and integers n & m, (xn)m = xnm.
Proof:
Fix n in Z.
Then since (xn)0 = e = x0 = x0m, the claim holds for m = 0.
Now assume the statement holds for some k . Then from (b),
(xn)k+1 = (xn)k xn = xnk xn = xnk+n = xn(k+1), so it holds for k + 1.
Therefore, by mathematical induction, it must hold for all m > 0.
Finally, let m < 0. Then (xn)m = ((xn)-1)-m = (x-n)-m = x(-n)(-m) = xnm.
Part (d) Let G be abelian. Then for any x & y in G, (x y)n = xn yn.
Proof:
Since (x y)0 = e = e e = x0 y0, the claim holds for n = 0.
Now assume that the statement holds for some k . Then
(x y)k+1 = (x y)k x y = (xk yk) x y = xk (yk x) y =(xk (x yk) y = (xk x) (yk y) = xk+1 yk+1,
and it holds for k + 1. Therefore, by the principle of induction, it holds for all n ≥ 0.
Finally, let n < 0. Then –n > 0 & (x y)n = ((x y)-1)-n = (x-1 y-1)-n = (x-1)-n (y-1)-n = xn yn.