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Let $S$ be a finite semigroup with $k$ elements. If we have at least $(nk)^k$ elements $s_1, \ldots, s_m$, i.e. $m \ge (nk)^k$, then the product $$ s_1 \cdot s_2 \cdots s_m $$ could be factored as $s_1 \cdot s_2 \cdots s_m = xy_1 y_2 \cdots y_n z$ (i.e. the elements $x,y_i,z$ combine a consecutive subsequence and equal their product) such that the product $y_i y_{i+1} y_{i+j}$ is the same for every permissible $i$ and $j$.

Does anyone has an idea how to solve this? As $y_1 = \ldots = y_n$ and $y_i y_{i+1} = y_i$ the element must be an idempotent, but I do not know how to choose the factorisation/decomposition?

StefanH
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    Isn't that equivalent to getting $s_1s_2\dots s_n=xw^nz$ with $w=w^2=\dots=w^n$ ? – Asinomás Aug 01 '16 at 19:20
  • @CarryonSmiling Yes, you are right. – StefanH Aug 01 '16 at 19:22
  • This is probably going to be a combination of induction and the pidgeon-hole principle. – Asinomás Aug 01 '16 at 19:25
  • related: http://math.stackexchange.com/questions/353028/is-there-an-idempotent-element-in-a-finite-semigroup – Asinomás Aug 01 '16 at 19:31
  • @CarryonSmiling I know this result that in a finite semigroup every element has an idempotent power, in particular $S$ contains at least one idempotent. But its application seems to be more tricky here... – StefanH Aug 01 '16 at 19:32
  • Isn't it enough to show that one of the consecutive subproducts is an idempotent? I don't see what the point of $n$ is. – Asinomás Aug 01 '16 at 19:37
  • The $y_i$ "combine the elements" from the sequence, i.e. we have $y_1 = s_{k_1}\cdots s_{k_2-1}, y_2 = s_{k_2}\cdots s_{k_3-1}$ and so on for some subsequence $(k_i)$, i.e. the $n$ says that we can group the elements in such a way, so in that sense your first comment is a little bit weaker and not equivalent which I noticed yet. If we can group the product in such a way what you wrote in your first comment would follow, but if this condition holds it it not said that we can group certain consecutive $s_i$'s as to give $n$ times $w$ from your comment. – StefanH Aug 01 '16 at 19:42
  • Oh yeah, I see your point. – Asinomás Aug 01 '16 at 19:44
  • :) it does not seem to be that straightforward, at least I do not see how to use induction and the pigeon-hole principle. Surely it is related to these principles, but I guess they must be applied in some very clever way... – StefanH Aug 01 '16 at 19:46

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