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In a group $G$, the non empty subsets form a semigroup (with identity) under the usual multiplication $ST=\{st \mid s\in S, t \in T\}$.

This semigroup seems to be very rich of information, for instance if $G$ is finite the idempotents in this semigroup are exactly the subgroups of $G$. Also for any (non empty) subset $S$, there is a power $S^n$ of $S$ which is an idempotent, and so a subgroup of $G$. if $S$ contains the identity, this would be the subgroup generated by $S$ (I don't know what happen if $S$ does not contain the identity).

1) Is there any attempts to studying such a semigroup, and its relations to the underlying group?

2) If $G$ is finite, our subsets can be identified to the elements of the group algebra $\mathbb{Z}_2[G]$, the multiplication in this algebra is somewhat different from the above one; can one covers the information about the semigroup of subsets of $G$ by only studying the group algebra $\mathbb{Z}_2[G]$?.

Thanks in advance.

Andreas Caranti
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Yassine Guerboussa
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  • I assumed that $G$ finite. In general, for any element $a$ in finite semigroup, there is some (positive) power of $a$ which is idempotent. – Yassine Guerboussa Oct 03 '13 at 10:12
  • Of course. I fail at reading. Again. – kahen Oct 03 '13 at 10:14
  • I also fail in reading in most times. – Yassine Guerboussa Oct 03 '13 at 10:16
  • Related: http://math.stackexchange.com/questions/101045/what-can-we-learn-about-a-group-by-studying-its-monoid-of-subsets and http://math.stackexchange.com/questions/475657/a-special-subgroup-of-groups-of-order-n/ – Mikko Korhonen Oct 03 '13 at 13:28
  • Yes, I think the first link covers my first question. Also I was thinking about defining the order of $S$ (by which I mean the smallest positive integer $n$ such that $S^n$ is idempotent (a subgroup)), Now I know from your second link that $n$ can not exceed the order of $G$, and perhaps it divides it!!. This semigroup seems to hide many surprises. Thank you Mikko Korhonen. – Yassine Guerboussa Oct 03 '13 at 14:02
  • @YassineGuerboussa: It turns out the "order" of $S$ does not necessarily divide $|G|$. For example, $S = {(1,2), (2,3,4)}$ (permutations from $S_4$) has order $7$ and $S = {(1,2)(3,4), (1,2,3)}$ has order $5$. – Mikko Korhonen Oct 03 '13 at 16:14
  • Or just the cyclic group $G = {1, g, g^2}$ of order $3$ and $S = {g, g^2}$ which satisfies $S^2 = G$. – Mikko Korhonen Oct 03 '13 at 18:09
  • Yes, I was much hopeful. – Yassine Guerboussa Oct 03 '13 at 18:20

1 Answers1

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The first question has been answered by a link in the comments. I will post it in an answer: What can we learn about a group by studying its monoid of subsets?

Indeed, as the second answer to that question says, these semigroups have been studied. They haven't been studied a lot because of the difficulty -- Lyapin's early book on semigroups has a paragraph on them in which the author dismisses the construction as too difficult to study.

I find it interesting that you thought about the group algebra since I thought about it too. The question the link directs to is actually mine (I'm still getting a newsletter despite having deleted my account), and you can see the thought process that lead me to thinking about it there. The power semigroup, as it is called, is isomorphic to a certain variation on the group algebra idea, with $\mathbb Z_2$ replaced by $(\{0,1\},\vee,\wedge)$ and the requirement of finite supports dropped. With finite supports you get the so called finitary power semigroup (the semigroup if finite subsets), which has also been studied. This gives the idea of thinking of the elements of $\mathbb Z_2[G]$ as subsets of $G$ with a certain funny ring (or algebra) structure. I did find it fascinating when I noticed it and asked a certain professor about it, but my excitement was quelled by the professor's indifference. Power semigroups are pretty much the only mathematical structures I work on at the moment (it started with the linked question), but I have never seen any mention of even the similarity of the power semigroup construction to the group algebra construction in the literature.

For the question on orders in the comment, you might find this question interesting as it deals with exactly that in $\mathbb N$.

Community
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ymar
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    If k is any semiring then one can define the group algebra kG. For a finite group G, P(G) is the group algebra of G over the Boolean semiring. – Benjamin Steinberg Oct 10 '13 at 01:31
  • @user23211 : That is an excellent answer. I wish you a good luck in your studies. (Perhaps it is better to examine your ideas first with some straightforward or small groups (with computer if needed)) Best wishes. – Yassine Guerboussa Oct 15 '13 at 20:55