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Refer to Lang's Algebra p. 80 exercise 46. Let $G$ be a finite group acting on a finite set $S$. Then Lang calls a partition $S = \bigcup_{i \in I} S_i$ of $S$ "stable" if $G$ maps each $S_i$ onto some $S_j$.

Does that mean that for any $i \in I$ and for any $x \in G$ there exists $j \in I$ such that $x S_i=S_j$ or that for any $i \in I$ there exists $j \in I$ such that $x S_i = S_j$ for all $x \in G$? Notice that in the first interpretation $j$ depends on $x$, while in the second interpretation it does not. Which of the two is the correct interpretation?


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Define "precise", please. – Henning Makholm Sep 30 '11 at 18:50
Could you be more specific? I don't see anything imprecise about it, offhand. – Dylan Moreland Sep 30 '11 at 18:51
By precise, i mean, is it correct? – Manos Sep 30 '11 at 19:04
Just to clarify, this is not my own definition. "Stable" partition is already existing terminology in the theory of primitive groups. – Manos Sep 30 '11 at 19:31
After some thought, it seems more reasonable that $j$ should not depend on $x$. I.e. the definition would be that for any $i \in I$ there exists $j \in I$ such that for all $x \in G$ we have $x S_i = S_j$. – Manos Sep 30 '11 at 19:43
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I think "stable partition" means whatever Lang chose it to mean, i.e., it's ambiguous. Note that if $xS_i=S_j$ for all $x$ then $eS_i=S_j$ for the identity, and so $i=j$.

My assumption would be that he intended that each element $x$ of $G$ maps each cell $S_i$ to a cell $S_j$ where $j$ depends on $x$. This is what happens if the cells of the partition are the blocks of imprimitivity of a transitive group $G$. In the permutation group literature this would more likely be described as a $G$-invariant partition.

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That's where i concluded as well. Thanks. – Manos Oct 1 '11 at 3:13

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