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Prof. Pinter's "A Book of Abstract Algebra" presents this exercise:

Let $G$ be a group and $a\in G$

Prove that $a^{m}a^{n}=a^{m+n}$ where $m < 0$ and $n > 0$

To prove $a^{m}a^{n}=a^{m+n}$ where $m < 0$ and $n > 0$, I simply wrote out:

$$a^{-m}a^{n} = a^{-1} ... a^{1}...$$ where $...$ indicates the value of $m$ and $n$, respectively.

Since the $a^{-1}$ will cancel out with the $a^{1}$ since $a^{-1}a^{1}=\epsilon$, then it seems to me that $a^{m}a^{n}=a^{m+n}$ where $m < 0$ and $n > 0$.

This doesn't seem to be technical enough to be a proof. Please tell me it's sufficient.

Community
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Kevin Meredith
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    It certainly looks like it's on the right lines, although you could clean up the presentation of the multiplied out version eg: $$\underbrace{a^{-1}...a^{-1}}{|m|\text{ times}}\underbrace{a...a}{n\text{ times}}$$ and maybe discuss the different outcomes if $|m|$ or $n$ is larger. – Joffan Jun 04 '15 at 20:16

2 Answers2

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First show that $a^m = (a^{-1})^{-m}$ for $m<0$ by induction. Clearly, $a^{-1} = (a^{-1})^{1}$, and if $a^m = (a^{-1})^m$ for some $m<0$, then $$a^{m-1} = a^ma^{-1} = (a^{-1})^{-m}a^{-1}=(a^{-1})^{-(m-1)}. $$

Then $-m>0$, so as $a^{-1}a^n = a^{n-1}$, by a similar induction we have $$a^ma^n = (a^{-1})^{-m}a^n = a^{n-(-m)}=a^{n+m}. $$

Math1000
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Let's prove it for any $m$ and $n\ge0$ by induction on $n$. The statement is obvious for $n=0$. So assume it holds for $n$: $$ a^ma^{n+1}=a^m(a^na)=(a^ma^n)a=a^{m+n}a $$ Now we're reduced to prove that, for any $k$, $a^ka=a^{k+1}$.

This is true for $k\ge0$, by definition of powers, so assume $k<0$. Then $a^k=(a^{-1})^{-k}=(a^{-1})^{-k-1}a^{-1}$ (again by definition of powers. Then $$ a^ka=(a^{-1})^{-k-1}a^{-1}a=(a^{-1})^{-k-1}=a^{k+1} $$

egreg
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