Possible Duplicate:
A cyclic subsemigroup of a semigroup S that is a group
My homework: An element $s^{i+k}$ on the cycle is idempotent iff
$$ s^{i+k} = s^{2i+2k} ,$$
or equivalently
$$ i+k = 2i+2k \pmod p .$$
I'm stuck here (this is my first modulo equation).
Also, is there algebraic proof, not referring to semigroup structure depicted on Figure 1?
That last comment of mine is long enough to turn into a full-fledged hint:
Let $t = s^{i+k}$; $t$ is idempotent iff $t=t^2$, i.e., iff $s^{i+k} = (s^{i+k})^2 = s^{2i+2k}$, so the real problem is showing that this is equivalent to the statement that $i+k \equiv 2i+2k \pmod p$, i.e., that $p$ divides $(2i+2k)-(i+k)=i+k$. One way to calculate $s^{2i+2k}$ is to start with $s^{i+k}$ and multiply by $s$ $i+k$ times. Each multiplication by $s$ advances the result one step around the cycle, so going from $s^{i+k}$ to $s^{2i+2k}$ takes $i+k$ steps. Clearly $s^{2i+2k}=s^{i+k}$ iff you end up back where you started, so $i+k$ steps take your from $s^{i+k}$ back to $s^{i+k}$. In other words, you must have gone round the cycle a whole number of times. What does that tell you about $i+k$?