# Subsemigroup generated by an element contains unique idempotent [duplicate]

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?

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## marked as duplicate by Mariano Suárez-Alvarez♦Apr 3 '12 at 2:26

$a \equiv b \pmod p$ just means that $p$ is a divisor of $a-b$, so you’re being asked to show that $s^{i+k}$ is idempotent iff $p$, the cycle length, is a divisor of $(2i+2k)-(i+k)=i+k$. Does that help at all? – Brian M. Scott Sep 26 '11 at 21:18
Thanks Brian, it reduced to proving that there exists unique $k$ such that $p$ is divisor of $i+k$. In other words, for any natural numbers $p$ and $i$ there exist $t$ and $k$ such that $pt = i+k$... I'm having doubts about validity of $i+k = 2i+2k \pmod p$ – Tegiri Nenashi Sep 26 '11 at 21:40
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? – Brian M. Scott Sep 26 '11 at 21:49

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 $i+k$?
From $s^{i+k} = s^{2i+2k}$ it follows that $i+k$ is "i-p attractor equivalent" to $2i+2k$, where "i-p attractor" is a chain of length i attached to a cycle of length p. This is the same as $(i+k)-i = (2i+2k)-i (mod p)$. Does it follow $i+k = 2i+2k (mod p)$ ? – Tegiri Nenashi Sep 26 '11 at 22:35
$(i+k)-i\equiv(2i+2k)-i\pmod p$ means that $p$ divides $[(2i+2k)-i]-[(i+k)-i]$, and $i+k\equiv 2i+2k\pmod p$ means that $p$ divides $(2i+2k)-(i+k)$; are those statements equivalent? – Brian M. Scott Sep 26 '11 at 22:43
Wasn't thinking clearly:-( $a=b (mod p) \Leftrightarrow a+c=b+c (mod p)$ So, where are we now? For any natural numbers $p$ and $i$ there exist $t$ and $k$ such that $pt = i+k$... The only Diophantine equation I'm familiar with is $ap-bq=1$. – Tegiri Nenashi Sep 26 '11 at 22:56
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. – Brian M. Scott Sep 26 '11 at 23:14