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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?

Thanks.

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    After some more thought, it seems to me that $j$ should actually depend on both $x$ and $i$. Otherwise, for any cell $S_i$ of the partition we have $e S_i = S_i$ and so that would imply that $x S_i = S_i$ for all $x \in G$, which is not general enough.2011-09-30

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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.2011-10-01