2
$\begingroup$

Possible Duplicate:
A cyclic subsemigroup of a semigroup S that is a group

Subsemigroup generated by an element contains unique idempotent

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?

  • 0
    But $k$ can’t be unique: once you find one $k$ s.t. $p$ divides $i+k$, you can keep adding $p$ to it to get infinitely many more. Think about it this way. Let $t=s^{i+k}$; $t$ is idempotent iff $t^2=t$. But $t^2=(s^{i+k})^2=s^{2i+2k}$, $t$ is idempotent iff $s^{i+k}=s^{2i+2k}$. Now start with $s^{i+k}$; every time you multiply by $s$, you go one step forward around the cycle. After $i+k$ steps you want to be back where you started, so you have to have gone round the cycle a whole number of times. Therefore ... what?2011-09-26

1 Answers 1

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 $i+k$?

  • 0
    You’re making this *way* too hard. For $i+k$ steps to get you back to $s^{i+k}$, $i+k$ has to be a multiple of $p$, right? And if it’s a multiple of $p$, then $p$ is a divisor of $(2i+2k)-(i+k)$. You don’t need to know *what* multiple of $p$ it is.2011-09-26