Group Actions and Normal Subgroups

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SUMMARY

The discussion focuses on the action of a group G on the set of left cosets X of a subgroup H. It establishes that the formula g(xH) = gxH defines a group action. Furthermore, it concludes that H is a normal subgroup of G if and only if every orbit of the induced action of H on X contains a single point, which is demonstrated by showing that left and right cosets coincide when H is normal.

PREREQUISITES
  • Understanding of group theory concepts, specifically group actions.
  • Familiarity with left cosets and their properties.
  • Knowledge of normal subgroups and their definitions.
  • Proficiency in mathematical proofs and logical reasoning.
NEXT STEPS
  • Study the definition and properties of group actions in detail.
  • Explore the relationship between normal subgroups and cosets in group theory.
  • Learn about the implications of orbits and stabilizers in group actions.
  • Investigate examples of normal subgroups in various groups, such as symmetric groups.
USEFUL FOR

Mathematics students, particularly those studying abstract algebra, group theorists, and anyone interested in the properties of normal subgroups and group actions.

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Homework Statement


Let H be a subgroup of G and let X be the set of left cosets of H in G.
Show that the formula g(xH) = gxH is an action of G on X.
Prove that H is a normal subgroup of G if and only if every orbit of the induced action of H on X contains just one point.


Homework Equations





The Attempt at a Solution


I've shown that the formula defines an action.
For the second part, since H is normal its left and right cosets are equivalent. Thus we consider h(xH) for h in H. So h(xH) = hxH = Hhx. But since h is in H this is the same as Hx. So for all h, the orbit contains only the point Hx.
However, the converse of the second part is what is giving me trouble. We know that each orbit contains only one element, but I'm not sure what else we can can from that.
 
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H is a subgroup, so it includes e. For all h, h(xH)=e(xH)=xH. Left-multiplying by xi (x inverse) on both sides gives xihxH=H, so xihx belongs to H (for all x in G and h in H).
 

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