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Mathematics
Linear and Abstract Algebra
Using group action to prove a set is a subgroup
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[QUOTE="fresh_42, post: 6029447, member: 572553"] True. I also would have considered the direct way easier to see. The point of this exercise is to prepare for the - now I don't know the exact English term - we call it orbit formula. It is of similar importance as Lagrange's theorem for finite groups. It says: Let ##(G,\cdot )## be a group which operates on a set ##M##, say ##(g,m) \longmapsto g.m \in M\,.## Then we have for every ##m \in M## a bijection $$G/G_m \longleftrightarrow G.m = \{\,g.m\, : \,g \in G\,\} \subseteq M$$ where ##G_m## is the stabilizer and ##G.m## the orbit of ##m## under the operation of ##G##. Note that ##|G/G_m|## isn't necessarily a group, as the stabilizer may not be a normal subgroup. However, one can still build the equivalences classes ##g \,\cdot \,G_m##, it's just not necessarily a group again. This yields for finite sets and groups ##|G/G_m| = |G.m|## resp. ##|G| = |G_m|\,\cdot \,|G.m|\,.## [/QUOTE]
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Using group action to prove a set is a subgroup
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