Proving Basic Exponent Properties for a Group

[middleCmusic]

When proving that x^m x^n = x^{m+n} and that (x^m)^n = x^{mn} for all elements x in a group, it's easy enough to show that they hold for all m \in \mathbb{Z} and for all n \in \mathbb{N} using induction on n. The case n = 0 is also very easy. But how does one prove this for n \in \mathbb{Z}^{-}?

I tried to do it by using the fact that n = - \nu for some \nu \in \mathbb{N}, but this didn't get me anywhere. Do you have to do induction on the negative integers separately? I'm sure there's a simple answer to this question that I'm just not seeing.

Note that I'm working with the standard recursive definition of exponents, and the definition x^{-n} = (x^{-1})^n.

 

[middleCmusic]

Update: I think I figured it out. I'd still appreciate a simpler way, if anyone has one.

For a negative integer n, set n = - \nu as before.
Then
<br /> \begin{align}<br /> x^m x^n <br /> &amp;= x^m x^{- \nu} \\<br /> &amp;= x^m (x^{-1})^{\nu} \\<br /> &amp;= ((x^{-1})^{-1})^m (x^{-1})^{\nu} \tag*{since $x = (x^{-1})^{-1}$} \\<br /> &amp;= (x^{-1})^{-m} (x^{-1})^{\nu} \tag*{by the definition of negative exponents} \\<br /> &amp;= (x^{-1})^{-m+\nu} \tag*{by the rule for positive exponents}\\<br /> &amp;= (x^{-1})^{-(m+ (-\nu))}\\<br /> &amp;= (x^{-1})^{-(m+n)} \\<br /> &amp;= ((x^{-1})^{-1})^{m+n} \\<br /> &amp;= x^{m+n} \tag*{since $x = (x^{-1})^{-1}$}<br /> \end{align}<br />

 

[Office_Shredder]

If you liked induction for positive n you can do induction for negative n also. Show if it holds for n, then it holds for n-1 as well