for any elements a and b from a group and any integer n, prove that (a^-1ba)^n=a^-1b^na

Question

This is from abstract algebra and it is a proof by induction

Answer 1

Yes, this can be proven by induction:

 

Foundation step: n=1

When n=1, the equality is obvious: (a^-1ba)¹ = a^-1ba = a^-1b¹a

 

Induction step: suppose that for n=k this equality is correct, let's prove that it is also correct for n=k+1:

(a^-1ba)^k+1 =(a^-1ba)^k(a^-1ba)

                =(a^-1b^ka)(a^-1ba)  use the equality when n=k

                =a^-1b^kaa^-1ba

                =a^-1b^kba

                =a^-1b^k+1a

So the equality when n=k+1 is proven to be correct.

 

By induction, we can conclude that for all integers n>=1, the equality is correct.

 

 

 

Answer 2

In a proof by induction you have 3 steps.

 

1. Prove P(1) true

 

2. Assume P(k) true

 

3. Prove P(k+1) true

 

In your question when you say any integer n  for simplicity let's start with the positive integers.

 

1. for n=1 P(1): a^-1b¹a = (a^-1ba)¹   which is true

 

2. assume P(k): a^-1b^ka = (a^-1ba)^k  is true

 

3. prove P(k+1): a^-1b^k+1a = (a^-1ba)^k+1  is true

 

Go from (2) to (3) using the associative property of the group.

 

 

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