Proving A5 has No Normal Subgroups: Conjugacy Classes Approach

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A5 has no normal subgroups other than itself and the trivial group {e}. The discussion focuses on using conjugacy classes to prove this, as normal subgroups must be unions of conjugate classes. The main challenge is determining the conjugacy classes of A5. The thread suggests referencing the conjugacy classes of elements in S5 to aid in this process. Understanding these classes is crucial for completing the proof.
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?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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