Proving A5 has No Normal Subgroups: Conjugacy Classes Approach

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SUMMARY

The discussion focuses on proving that the alternating group A5 has no normal subgroups other than itself and the trivial subgroup {e}. The approach suggested involves analyzing the conjugacy classes of A5, as a normal subgroup must be a union of these classes. The user seeks guidance on determining the conjugacy classes, with a reference to the conjugacy classes of elements in the symmetric group S5 as a potential starting point.

PREREQUISITES
  • Understanding of group theory, specifically the properties of normal subgroups.
  • Familiarity with the concepts of conjugacy classes in group theory.
  • Knowledge of the structure and properties of the alternating group A5.
  • Basic understanding of the symmetric group S5 and its relation to A5.
NEXT STEPS
  • Research the conjugacy classes of the symmetric group S5.
  • Study the properties of the alternating group A5 in detail.
  • Learn about centralizers and their role in group theory.
  • Explore proofs regarding normal subgroups in finite groups.
USEFUL FOR

Mathematicians, particularly those studying group theory, graduate students in abstract algebra, and anyone interested in the properties of the alternating group A5.

Obraz35
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Homework Statement


I am interested in proving that A5 has no normal subgroups except itself and {e}.


The Attempt at a Solution


Some proofs that I have seen use centralizers to do this, but since I haven't gone through that yet I think there should be some say to do it without them.

My approach would be to find the conjugacy classes of A5 and use their orders to show that there cannot be a normal subgroup in A5 since a normal subgroup is a union of conjugate classes.
But my main problem is how I should go about finding the conjugacy classes.

Thanks for your help.
 
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There is an identical thread to this in this forum. The advice there is: do you know the conjugacy classes of elements in S_5?
 

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