Embed Symmetric Group in Alternating Group A(n+2)

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SUMMARY

This discussion focuses on embedding the symmetric group S_n into the alternating group A_(n+2). The user proposes a method to define a map f that acts on a set U of n+2 symbols, where f(p) = p * t for odd permutations p in S_n, and f(p) = p for even permutations. This construction ensures that f is a homomorphism with a trivial kernel, confirming that A_n is indeed a subgroup of A_(n+2). The explicit mapping demonstrates the relationship between these groups effectively.

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  • Understanding of group theory, specifically symmetric and alternating groups.
  • Familiarity with homomorphisms and their properties in abstract algebra.
  • Knowledge of permutation notation and operations.
  • Basic comprehension of even and odd permutations.
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  • Study the properties of symmetric groups S_n and alternating groups A_n.
  • Learn about homomorphisms in group theory and their applications.
  • Explore explicit constructions of group embeddings in algebra.
  • Investigate the implications of kernel properties in group homomorphisms.
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Mathematicians, particularly those specializing in group theory, algebra students, and researchers exploring group embeddings and homomorphisms.

shaggymoods
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How would you embed the symmetric group on n letters in the alternating group of (n+2) letters? I'm actually trying to write down an explicit map but can't seem to come up with one. I know An will be a subgroup of A(n+2) but I have a feeling that a map that is the identity on An and not-the-identity elsewhere won't work. Any thoughts?
 
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Imagine a set U of n+2 symbols. A permutation p in Sn acts on the first n of these. If p is an odd permutation, define a new permutation f(p) on U by f(p) = p * t, where t is the extra transposition (n+1, n+2). Clearly, f(p) is in An+2. If p is an even permutation, define f(p) = p. It's easy to show that f is a homomorphism with trivial kernel.
 
Thanks!
 

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