To prove that $a^m.a^n=a^{m+n}$

Question

In a group $G$, I want to prove this theorem:

$$\forall a\in G,\; a^m.a^n=a^{m+n},\;m,n\in \Bbb Z.$$

I am thinking that only associative law is sufficient to prove this.please give some suggestions or hint to prove this.thanks in advance

Answer 1

It is clear that $a^{m+1}=a^m.a$. On the other hand, if $a^{m+n}=a^m.a^n$, then$$a^{m+n+1}=a^{m+n}.a=a^m.a^n.a=a^m.a^{n+1}.$$

Answer 2

Write out what each exponent means. We have $$\begin{align} a^ma^n&=\underbrace{a\times \dots \times a}_{m\text{ times}}\times\underbrace{a\times \dots \times a}_{n\text{ times}} \\ &=\underbrace{a\times \dots \times a}_{m+n\text{ times}} \\ &=a^{m+n}. \end{align}$$

Yes, associativity is thus sufficient.