Abstract Algebra: Commutative Subgroup

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SUMMARY

The discussion centers on proving that the set H = {x in G | axb = bxa} forms a subgroup of a group G, where a and b are elements that commute. The user employs the subgroup test, successfully demonstrating the existence of the identity element and closure under multiplication. The challenge lies in proving that the inverse of any element x in H also belongs to H. A suggested approach involves using the commutation relation and the properties of inverses to establish the necessary condition.

PREREQUISITES
  • Understanding of group theory concepts, specifically subgroups.
  • Familiarity with the subgroup test in abstract algebra.
  • Knowledge of commutative properties in group operations.
  • Ability to manipulate algebraic expressions involving group elements and their inverses.
NEXT STEPS
  • Study the subgroup test in detail, focusing on its application in various group structures.
  • Learn about the properties of commutative groups and their implications for subgroup formation.
  • Explore the concept of inverses in group theory and how they relate to subgroup criteria.
  • Investigate examples of subgroups in specific groups, such as symmetric groups or cyclic groups.
USEFUL FOR

This discussion is beneficial for students of abstract algebra, particularly those studying group theory, as well as educators seeking to clarify subgroup properties and proofs.

pat bismark
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Homework Statement



Let G be a group and let a, b be two fixed elements which commute with each other (ab = ba). Let H = {x in G | axb = bxa}. Prove that H is a subgroup of G.


Homework Equations



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The Attempt at a Solution



I'm using the subgroup test. I know how to show that the identity of G exists in H and that if x1, x2 exist in H then x1 times x2 exists in H but need help proving that if x is in H then x^-1 is also in H. I've tried starting with axb = bxa and multiplying on the right and left with various combinations of a, x^-1, and b, but can't get it to the form ax^-1b = bx^-1a.

Please let me know if you have any ideas, thanks.
 
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Prove that [b-1,a] = [b,a-1] = 0 and use that along with the fact that axb = bxa.
 

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