# Show that $a^{m} a^{n} = a^{m+n}$

Prove that if $G$ is a group and $a\in G$, then we have $\forall m,n\in\mathbb{Z}$ that $$a^{m} a^{n} = a^{m+n}.$$

I've proved the case when $m,n>0$ but I'm stuck on how to prove the case when one or both are negative without assuming that $(a^{-1})^n = a^{-n}$.

• How about using $a^n(a^{-1})^n=(aa^{-1})^n=1$? Nov 1, 2015 at 21:32
• If $n$ is negative you have to define $a^n=(a^{-1})^{|n|}$, than you can show that $(a^{-1})^{|n|}=(a^{|n|})^{-1}$. Nov 1, 2015 at 21:49
• @EmilioNovati You could also define $a^n = (a^{\left| n \right|})^{-1}$...
– A.P.
Nov 1, 2015 at 21:57
• Yes. I agree. The point is that the two definitions are equivalent. Nov 1, 2015 at 22:01

I assume that you would like to use $(a^{-1})^n = a^{-n}$ by saying that if $n > 0$ then $$a^m a^{-n} = a^m (a^{-1})^n = a^{m-n}$$ by leveraging the case for $m,n > 0$. The problem is that you can't really do this, because in that case you had two powers with the same basis, but usually $a \neq a^{-1}$.

Instead, what you could go this way. First observe that, up to swapping indices and "inverting everything", we can assume $m \geq 0$ and $m \geq n$.

Now, you already proved the case with $m,n$ positive, so let's consider $n \leq 0$. If $n$ or $m$ are $0$ the claim is clearly true (because $a^0 = e$). Otherwise assume $m > 0$ and suppose the claim true for $n$. To prove it for $n - 1$ observe that $$a^m a^{n-1} = (a^{m-1} a) (a^{-1} a^n) = a^{m-1} e a^n = a^{m-1+n}.$$

You have to consider a few cases to prove this theorem. In that way, I find it a bit annoying. Perhaps there is a better way, but I do not know of one.

Edit: There actually is a straightforward way of shortening this; see A.P.'s comment.

You have considered one such case ($m,n > 0$). Here we will prove (most of) the others. We will call case (i) the one you proved and begin with case (ii). The identity will be denoted $e$.

As for your question regarding $(a^{-1})^n=a^{-n}$, we have that $(a^{-1})^n=(a^n)^{-1}$ and that is, by definition of the notation, equal to $a^{-n}$.

Now, let us complete the proof by considering the other cases.

(ii) Suppose $n=0$, then $a^{m+n}=a^{m+0}=a^m$ and $a^m a^n=a^m a^0 = a^m e = a^m$, hence $a^{m+n}=a^m a^n$.

(iii) Suppose $n<0, m=-n$, then $a^{m+n}=a^0=e$, hence $a^m=a^ma^{-m}=e$ (this too is worth proving), so $a^m a^n = a^{m+n}$.

(iv) Suppose $n<0, m<-n$, then

\begin{aligned} a^m a^n & = a^m (a^{-1})^{-n} \\ & = a^m (a^{-1})^{m-(m+n)} \\ & = a^m \left((a^{-1})^m (a^{-1})^{-(m+n)} \right) \\ & = a^m \left( a^{-m}(a^{-1})^{-(m+n)}\right) \\ & = (a^m a^{-m}) (a^{-1})^{-(m+n)} \\ & = e (a^{-1})^{-(m+n)} \\ & = a^{m+n} \end{aligned}

(v) Suppose $n<0, m>-n$, then

\begin{aligned} a^m a^n & = a^{(m+n)+(-n)}a^n \\ & = (a^{m+n}a^{-n})a^n \\ &= a^{m+n}(a^{-n}a^n) \\ &= a^{m+n} e \\ &= a^{m+n} \end{aligned}

Thus far we have proven that for a fixed $m>0$ we have the identity for all $n$, whether $n<0$, $n=0$, or $n>0$. To be complete we would have to show the same for fixed $m<0$ and $m=0$. For $m=0$ we have an easy verification; for $m<0$ one can go through a similar argument as above. These cases will be left to the reader, mainly because I am too lazy to write it all out.

Hence, we have proved that if $a \in G$, then for all $m,n \in \mathbb{Z}$

$$a^m a^n = a^{m+n}$$

$\blacksquare$

• For $m<0$ just observe that you can reduce to one of the previous cases as $$a^{m+n} = (a^{-m-n})^{-1} = (a^{-m}a^{-n})^{-1} = a^m a^n.$$ You can also get rid of a couple of other cases by leveraging the fact that $a$ commutes with itself.
– A.P.
Nov 1, 2015 at 23:03

"without assuming that $(a^{-1})^n=a^{−n}$".

How was $a^{-n}$ (n positive) originally defined (if it was)? As notation it doesn't have any intrinsic meaning. It's very easy to show $(a^{-1})^n = (a^n)^{-1}$ (multiply either by $a^n$ and the result is $e$ so both of them are the unique inverse of $a^n$). So it's reasonable and well defined to define $a^{-n}$ as either one.

But while you're at it, you also have to define $a^0 = e$.