SUBSET K of elements in a group with finite distinct distinct conjugates

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SUMMARY

The discussion focuses on proving that a set K, consisting of elements with finite distinct conjugates, forms a subgroup under group operations. The participants emphasize the necessity of demonstrating closure under the group operation and the inverse property. The key argument presented is that if elements a and b in K have finite distinct conjugates, then the product ab must also have finite distinct conjugates, thus ensuring that ab remains in K. The clarification that the set of all conjugates of an element is finite, rather than the conjugates themselves, is crucial to the proof.

PREREQUISITES
  • Understanding of group theory concepts, specifically subgroups and conjugates.
  • Familiarity with group operations and their properties.
  • Knowledge of finite groups and their characteristics.
  • Basic proficiency in mathematical proofs and logical reasoning.
NEXT STEPS
  • Study the properties of conjugacy classes in group theory.
  • Learn about the criteria for subgroup formation in finite groups.
  • Explore examples of groups with finite conjugates, such as symmetric groups.
  • Investigate the implications of conjugacy in the context of group homomorphisms.
USEFUL FOR

This discussion is beneficial for mathematicians, particularly those specializing in abstract algebra, as well as students studying group theory who seek to deepen their understanding of subgroup properties and conjugacy relations.

Bachelier
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WTS, is that such set is a subgroup.

I need to show closure under group operation and inverse.

I can do the inverse which is usually the hardest part, but I'm stuck on the grp op.

So let a in K and b in K, both have finite distinct conjugates. Their conjugates are in the group too. WTS is that ab in K too.

if a, b in K then xax-1 = c in K and yay-1

consider xy(ab)(xy)-1 note since xax-1 and yay-1 are finite and distinct then x and y are finite and distinct hence xy and its inverse is finite

hence xy(ab)(xy)-1 in K

what do you think?
 
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it looks like you're confusing elements and sets.

the SET of all conjugates of a {xax-1: x in G} is finite, not the conjugates themselves.

so you want to show that if the set of conjugates of a is finite, and the set of conjugates of b is finite, so is the set of conjugates of ab.

it might be helpful to note that if x is ANY element of G:

x(ab)x-1 = (xax-1)(xbx-1).
 

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