Alternating group is the unique subgroup of index 2 in Sn?

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SUMMARY

The discussion confirms that the alternating group \( A_n \) is the unique subgroup of index 2 in the symmetric group \( S_n \) for \( n \geq 2 \). It establishes that any subgroup \( K \) of index 2 is normal and must be a union of conjugacy classes. The relationship between the sizes of \( A_n \), \( K \), and their intersection is explored, leading to the conclusion that \( |A_n \cap K| \) must be at least \( \frac{1}{4}n! \), reinforcing the uniqueness of \( A_n \) as the only subgroup of this index.

PREREQUISITES
  • Understanding of group theory concepts, specifically normal subgroups.
  • Familiarity with symmetric groups \( S_n \) and alternating groups \( A_n \).
  • Knowledge of conjugacy classes in group theory.
  • Basic combinatorial principles related to group orders and indices.
NEXT STEPS
  • Study the properties of normal subgroups in group theory.
  • Research the structure and properties of symmetric groups \( S_n \) and alternating groups \( A_n \).
  • Learn about conjugacy classes and their significance in group theory.
  • Explore the concept of simple groups and their implications in subgroup structures.
USEFUL FOR

Mathematicians, particularly those specializing in abstract algebra, group theorists, and students studying the properties of symmetric and alternating groups.

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Where n >= 2.

Is this true or false? I only got so far:

If K is a subgroup of index 2, then it's normal. K is normal in Sn, so it's a union of conjugacy classes. Also, since |An K| = |An| |K| / |An intersection K| = 1/2n! * 1/2n! / |An intersection K| <= n!, then 1/4n! <= |An intersection K|.

I don't know how to come up with a counter-example or proof from there. Could anybody help?
 
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An is simple. Try using that.
 

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