How could I prove that the equality $a^na^m=a^{n+m}$ in a semigroup with identity?

Question

Let $S$ be a semigroup with a identity. How could I prove that the equality $a^na^m=a^{n+m}$ which holds for all $n,m \in \mathbb Z$. Note that $a \in S$ and $\mathbb Z$ denotes the set of all integers.

thanks for your help.

Answer 1

Consider 4 cases:

$m\ge 0, n\ge 0,$

$m\ge 0, n\le 0,$

$m\le 0, n\ge 0,$

$m\le 0, n\le 0.$

Have some difficulties?

Answer 2

for the first case m≥0,n≥0, (a^n)(a^m)= aaa..a(m factor) * aaa...a (n factor)= aaa...a(m+n factor )= a^(m+n)

for the second and third use mathmatical induction and for the fourth use as what i use for the first but by -m and -n factor .