- #1

Bashyboy

- 1,421

- 5

## Homework Statement

Let ##x \in G## and ##a,b \in \mathbb{Z}^+## Prove that ##x^{a+b} = x^a x^b##.

## Homework Equations

## The Attempt at a Solution

If I am not mistaken, we would have to do multiple induction on ##a## and ##b## for the statement/proposition ##P(a,b) : x^{a+b} = x^a x^b##. First we would have to show that ##P(a,1)## and ##P(1,b)## are true (base cases). But due to symmetry, it suffices to prove ##P(a,1): x^{a+1} = x^a x##. This is where I had trouble; I thought I would have to do induction on the statement on the statement ##P(a,1)##.

After trying many things, I decided to consult this https://crazyproject.wordpress.com/2010/01/04/laws-of-exponents-in-a-group/ This person appears to be doing the same as myself. But I found it unsettling that no justification was given for ##x^{a+1} = x^a x## (the base case); what is written is this: "For the base case, note that

The only way I could think of justifying it is as follows:

##x^{a+1} = \underbrace{x \cdot x \cdot ... \cdot x}_{(a+1)~terms}##

##= \underbrace{(x \cdot x \cdot ... \cdot x)}_{a ~terms} x## (associative property)

##= x^a x## (by definition of what ##x^a## means)

But if we can do that, then why not do similar splitting with ##a^{a+b}## , thereby avoiding all induction?