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Free idempotent generated semigroups and endomorphism monoids of free $G$-acts (1402.4042v2)

Published 17 Feb 2014 in math.GR

Abstract: The study of the free idempotent generated semigroup $\mathrm{IG}(E)$ over a biordered set $E$ began with the seminal work of Nambooripad in the 1970s and has seen a recent revival with a number of new approaches, both geometric and combinatorial. Here we study $\mathrm{IG}(E)$ in the case $E$ is the biordered set of a wreath product $G\wr \mathcal{T}_n$, where $G$ is a group and $\mathcal{T}_n$ is the full transformation monoid on $n$ elements. This wreath product is isomorphic to the endomorphism monoid of the free $G$-act $F_n(G)$ on $n$ generators, and this provides us with a convenient approach. We say that the rank of an element of $F_n(G)$ is the minimal number of (free) generators in its image. Let $\varepsilon=\varepsilon2\in F_n(G).$ For rather straightforward reasons it is known that if $\mathrm{rank}\,\varepsilon =n-1$ (respectively, $n$), then the maximal subgroup of $\mathrm{IG}(E)$ containing $\varepsilon$ is free (respectively, trivial). We show that if $\mathrm{rank}\,\varepsilon =r$ where $1\leq r\leq n-2$, then the maximal subgroup of $\mathrm{IG}(E)$ containing $\varepsilon$ is isomorphic to that in $F_n(G)$ and hence to $G\wr \mathcal{S}_r$, where $\mathcal{S}_r$ is the symmetric group on $r$ elements. We have previously shown this result in the case $ r=1$; however, for higher rank, a more sophisticated approach is needed. Our current proof subsumes the case $r=1$ and thus provides another approach to showing that any group occurs as the maximal subgroup of some $\mathrm{IG}(E)$. On the other hand, varying $r$ again and taking $G$ to be trivial, we obtain an alternative proof of the recent result of Gray and Ru\v{s}kuc for the biordered set of idempotents of $\mathcal{T}_n.$

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