Expert: Paul Klarreich - 2/28/2007

Question
Hi,
Can you please help me with the following question:
i think induction is one way to do it and i have to take care of three cases
when n=positive,n=negative and n=0

But i don't know how to do it would u please help!

The question is:
Let G be a group, let a,b in G, and let n be (not necessarliy positive) integer.
prove if a and b commute, then (ab)^n=(a)^n (b)^n

Answer
Questioner:   Karlo
Category:  Advanced Math
 
Subject:  Proof (groups)
Question:  Hi,
Can you please help me with the following question:
I think induction is one way to do it and I have to take care of three cases when n=positive,n=negative and n=0

But i don't know how to do it would u please help!

The question is:

Let G be a group, let a,b in G, and let n be (not necessarily positive) integer.
prove if a and b commute, then (ab)^n=(a)^n (b)^n
...........................................
Hi, Karlo,

I received a similar question, Feb 23.

From that earlier proof:  

To prove that in a group, where a1, a2, ... an are elements of a group, that
(a1a2.....an)^(-1)=(an)^n.....(a2)^-1(a1)^-1

This can be proved using mathematical induction.  If you don't have a lot of practice with that, the steps are:

A. Prove the theorem for a basic case, like n=1 or n=2.
B. Assume the theorem true for n=k.
C. Use that assumption to prove it for n=k+1.

Base case: Prove that  (ab)' = b'a'
[I'll just use a,b instead of a1,a2 here.   And I will write ' instead of -1 for the inverse, to save typing and make things easier to read.]

b'a'(ab) = b'(a'a)b  [associative]
= b'(e)b  [definition of inverse]
= b'b  [definition of unity]
= e [inverse]

Now assume the theorem for  n = k, which means:

(a1 a2 .. ak)' = ak' ... a2' a1'  [Assumption.]

Next prove the theorem for n = k+1

(a1 a2 .. ak ak+1)' = ak+1' ak' ... a2' a1'   << to be proved.

(a1 a2 .. ak ak+1)' = ((a1 a2 .. ak) ak+1)'  [Assoc.]
= ak+1' (a1 a2 .. ak)' [Base case]
= ak+1' ak' ... a2' a1'  [By assumption]
.........................................................
NOW YOUR QUESTION IS DIFFERENT, OF COURSE, since the above did not assume commutativity in the group.

We'll try positive 'n' first, and use M.I. (see above description)

Theorem:  (ab)^n = a^n b^n,  and  n > 0.
Base Case: n = 1, or  (ab)^1 = ab = a^1 b^1.  [Pretty simple, as most base cases are.]

Assume: (ab)^k = a^k b^k
To Prove: (ab)^k+1 = a^k+1 b^k+1

Proof:
(ab)^k+1 = (ab)^k (ab)   << exponent rules.
= a^k b^k a b            << use of assumption.
= a^k a b^k b            << You said commutative, right?
= a^k+1 b^k+1            << exponent rules.

OK, that's it for positive exponents.  For a zero exponent, it's rather easy, and I'll leave that to you.

Now what does  a^(-k) mean?  Remember, we are not necessarily talking about numbers, here.  I shall assume that  a^(-k) means (a^k)', where (as I did above) the prime symbol means 'inverse'.   

So perhaps we will start by proving this 'lemma':

Lemma I: (a1 a2 ... am)' = a1' a2' ... am'
[I.E. The inverse of a product is the product of the inverses.]

If this were not a commutative group, you would need the proof I sent to the other student, but here all you have to do is to note that:

(a1 a2 ... am)(a1' a2' ... am') =
a1 a2 ... am a1' a2' ... am' =
a1 a1' a2 a2' ... am am' = e e ... e = e^m = e.

And this corollary is easy, by just setting all the a1..am to be the same:

Lemma II: (a^m)' = (a')^m

.................. READY TO GO..............
Theorem:  (ab)^n=(a)^n (b)^n,  and  n < 0.
Then let  n = -m, where m>0 and we have the equivalent:

Theorem:  (ab)^-m = a^-m b^-m

(ab)^-m = ( (ab)^m )'   << definition of negative exponent.
= ((ab)')^m             << Lemma 2.
= (a' b')^m             << Lemma 1.
= (a')^m (b')^m         << First half of theorem proved earlier; m > 0.
= a^-m b^-m             << definition of negative exponent.

That's about it, I think.