Proving Basic Exponent Properties for a Group

1. May 3, 2013

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 $x^{-n} = (x^{-1})^n$.

Last edited: May 3, 2013
2. May 3, 2013

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
\begin{align} x^m x^n &= x^m x^{- \nu} \\ &= x^m (x^{-1})^{\nu} \\ &= ((x^{-1})^{-1})^m (x^{-1})^{\nu} \tag*{since x = (x^{-1})^{-1}} \\ &= (x^{-1})^{-m} (x^{-1})^{\nu} \tag*{by the definition of negative exponents} \\ &= (x^{-1})^{-m+\nu} \tag*{by the rule for positive exponents}\\ &= (x^{-1})^{-(m+ (-\nu))}\\ &= (x^{-1})^{-(m+n)} \\ &= ((x^{-1})^{-1})^{m+n} \\ &= x^{m+n} \tag*{since x = (x^{-1})^{-1}} \end{align}

Last edited: May 3, 2013
3. May 3, 2013

Office_Shredder

Staff Emeritus
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