Suppose G is a group, with a ∈ G. Prove that $a^ma^n=a^{m+n}$ for all integers m and n.
Let m>0 and n=0
Then $a^ma^n=a^ma^0=a^me=a^m=a^{m+0}=a^{m+n}$.
So, for m>0 and n=0, then $a^ma^n=a^{m+n}$.
We will now consider four separate cases because of the various possibilities of signs that can occur as powers.
Case 1: Consider m,n∈$\mathbb{Z^+}$.
[Basis Step]
Let n=1. Then, $a^ma^n=a^ma^1=a^ma=aa^{m+1-1}, by definition,
=a^{m+1}=a^{m+n}$.
[End of Basis Step]
[Inductive Step]
Suppose that for some integer k>0 we have $a^ma^k=a^{m+k}$.
So, $a^ma^{k+1}=a^ma^1a^{k+1-1}=a^ma^ka^1=a^{m+k}a$, by induction hypothesis and substitution,
$=a^1a^{m+k+1-1}=a^{m+k+1}$, by definition.
Thus, $a^ma^{k+1}=a^{m+k+1}$.
[End of Inductive Step]
Therefore, for m,n∈$\mathbb{Z}$, $a^ma^n=a^{m+n}$.
Case 2:
Consider m∈$\mathbb{Z-}$ and n∈$\mathbb{Z+}$
[Basis Step]
Let n=1. Then, $a^ma^n=a^ma^1=(a^{-1})^{|m|}a=((a^{-1})(a^{-1})^{|m|-1})a=((a^{-1})^{|m|-1})(a^{-1}))a=(a^{-1})^{|m|-1}((a^{-1})a)=(a^{-1})^{|m|-1}e=(a^{-1})^{|m|-1}=(a)^{-(|m|-1)}=a^{-|m|+1}=a^{m+1}=a^{m+n}$
So, for m>0 and n=1, $a^ma^n=a^{m+n}$.
[End of Basis Step]
[Inductive step]
Suppose that for some integer k>0 we have $a^ma^k=a^{m+k}$.
So, for $a^ma^{k+1}=a^maa^{k+1-1}=a^maa^k=(a^{-1})^{|m|}aa^k=((a^{-1})(a^{-1})^{|m|-1})aa^k=((a^{-1})^{|m|-1})(a^{-1}))aa^k=(a^{-1})^{|m|-1}((a^{-1})a)a^k=(a^{-1})^{|m|-1}ea^k=(a^{-1})^{|m|-1}a^k$
This is where I'm stuck. and I still have to do m>0,n<0 and m<0,n<0
First do induction on $n\ge 0$, for arbitrary $m\in \Bbb Z$.
$n=0$. We have $a^ma^0=a^me=a^m=a^{m+0}$.
$n\mapsto n+1$: Assume that $a^ma^n=a^{m+n}$ for all integers $m$. Then $$ a^{m}a^{n+1}=a^m(a^na)=(a^ma^n)a=(a^{m+n})a=a^{m+n+1}. $$ Hence we are done. Note that we used associativity (not "transitivity"). Finally, do the same induction on $-n$.