Proof of x^a Inverse in Arbitrary Multiplicative Group

[icantadd]

Homework Statement

x is in the arbitrary multiplicative group, and a,b are positive integers.
given that
$$x^{a+b} = x^ax^b$$ and $$(x^a)^b =x ^{ab}$$
show
that
$$(x^a)^-1 = x^{-a}$$

Homework Equations

na

The Attempt at a Solution

Induction:

I) (x)^{-1} = x ^{-1}
II) Assume $$(x^n)^{-1} = x^{-n}$$, to prove that $$(x^{n+1})^{-1} = x^{-(n+1)}$$.

$$(x^{n+1})^{-1}) = (x^nx)^{-1} = x^{-1}x^{-n} = x^{-n-1}$$

Is the last step justified?

 

[Hurkyl]

Is the last step justified?

Seeing how you didn't provide a justification, no.

More seriously, if you cannot see a rigorous reason why that last step should be true, then you definitely haven't written a valid proof.

 

[icantadd]

Seeing how you didn't provide a justification, no.

More seriously, if you cannot see a rigorous reason why that last step should be true, then you definitely haven't written a valid proof.

Actually, this is what gets me!

The text explicitly states the following
"$$x^{-1}x^{-1}x^{-1} \ldots x^{-1} \text{ n terms }$$"

It should follow from this that
$$x^{n} = x^{-1}x^{-(n-1)}$$

I can't see a rigorous road from this description, per se.