Expert: Paul Klarreich - 3/4/2007

Question
Hi,

Let G be a group, let a,b in G, and let n and m be (not necessarliy positive) integers.
prove
(i)if a and b commute, then (ab)^n=(a)^n (b)^n
which you already proved but imentioned it here coz we might to refer to it when doing the next proofs.
Can you please help me proving the following:

(ii)(a^n)^m=a^(mn)
(iii)(a^m)(a^n)=a^(m+n)

Thanks

Answer
Questioner:   Karlo
Category:  Calculus
 
Subject:  Proofs(exponents) related to groups
Question:  Hi,

Let G be a group, let a,b in G, and let n and m be (not necessarily positive) integers.
prove
(i)if a and b commute, then (ab)^n=(a)^n (b)^n
which you already proved but imentioned it here coz we might to refer to it when doing the next proofs.
Can you please help me proving the following:

(ii)(a^n)^m=a^(mn)
(iii)(a^m)(a^n)=a^(m+n)

Thanks
........................................
Hi, Karlo,

I think that the general outline of the proof for (1) will work for the others -- assume m,n > 0 first and prove it by induction.  Then assume one or the other or both is negative and work it out using one of the 'Lemmas' from the previous proof.

I am going to assume that this property is true, probably as a definition:

a^n a = a^(n+1)
.................................................
I think it is easiest to prove (iii) first, for positive m,n:
(iii) Theorem:  (a^m)(a^n) = a^(m+n), m,n > 0  

Induction on n:

Base case: (a^m)(a^1) = (a^m) a = a^(m+1)

Assume: (a^m)(a^k) = a^(m+k)
Prove: (a^m)(a^k+1) = a^(m+k+1)

Proof: (a^m)(a^k+1) = (a^m)(a^k) a = a^(m+k) a = a^(m+k+1)
..................................
Now it becomes easy to prove (ii), for positive m,n:

(ii) Theorem:  (a^n)^m=a^(mn), m,n > 0.

Induction on m:

Base: m = 1:  (a^n)^1=a^n = a^(1n)

Assume: m = k. (a^n)^k = a^(kn)
To prove: m = k + 1:  (a^n)^(k+1) = a^((k+1)n)
Proof: (a^n)^(k+1) = (a^n)^k (a^n) = a^(kn) a^n = a^(kn + n) = a^((k+1)n)
............................................

Now the issue is to handle negative exponents:

(iii) Theorem: (a^m)(a^n) = a^(m+n), m > 0, n < 0

Let n = -p, with  p > 0

(a^m)(a^n) = (a^m)(a^(-p)) = (a^m)(a^p)'

Now there are three cases:  

A. m > p.  Write m = p + k,  and  k = m - p

(a^m)(a^n) = a^p a^k (a^p)' = a^k = a^(m - p) = a^(m - n)

B. m < p.  Write p = m + k,  and k = p - m

(a^m)(a^n) = (a^m)(a^p)' = (a^m)(a^(m+k))' =
(a^m)((a^m)(a^k))'  = (a^m)(a^m)'(a^k)' =
(a^k)' = a^(-k) = a^(m - p) = a^(m + n)

[Should be an easier way, but that looks right.]

C. m = p.  

(a^m)(a^n) = (a^m)(a^p)' = (a^m)(a^m)' = e,
or a^0 = a^(m-p) = a^(m+n)

...................................
And to handle negative exponents for:

(ii) Theorem:  (a^n)^m = a^(mn), m > 0, n < 0.

Let n = -p, with p > 0.

(a^n)^m = (a^(-p))^m = (a^p)')^m = ((a')p)^m = (a')^mp = a^(-mp) = a^(mn)

You can handle the case of m < 0, n > 0  yourself.