Solving Symmetric Group Induction Proof: Hints for Double Induction

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SUMMARY

The discussion focuses on proving the relationship between the transposition \((i,i+1)\) and the permutation \((1,2,...,n)(i-1,i)(1,2,...,n)^{-1}\) within the symmetric group \(S_n\). Participants debated the necessity of double induction versus single induction for this proof. Ultimately, one contributor highlighted that applying mappings for specific cases \(j < i-2\), \(j > i-1\), and \(j \in \{i-2,i-1\}\) simplifies the proof, rendering the use of induction unnecessary. The key insight is recognizing that the transformation \(\sigma (x_1,x_2,...,x_n) \sigma^{-1} = (\sigma(x_1),...,\sigma(x_n))\) trivializes the problem.

PREREQUISITES
  • Understanding of symmetric groups, specifically \(S_n\)
  • Familiarity with permutation notation and operations
  • Knowledge of induction principles, particularly double induction
  • Experience with group theory concepts and mappings
NEXT STEPS
  • Study the properties of symmetric groups and their elements
  • Learn about permutation mappings and their applications in group theory
  • Research induction techniques, focusing on when to use single versus double induction
  • Explore explicit formulas in group theory to understand their implications
USEFUL FOR

Mathematics students, particularly those studying abstract algebra, and anyone interested in group theory and induction proofs will benefit from this discussion.

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Homework Statement


Consider the symmetric group ##S_n##. I am trying to establish that ##(i,i+1)=(1,2,...,n)(i-1,i)(1,2,...,n)^{-1}##

Homework Equations

The Attempt at a Solution



I am trying to decide whether I need double induction or not. I have done several calculations to see whether I can get away with one induction, but it isn't clear to me whether this is possible. I could use some hints.
 
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Why don't you simply apply the mappings for the cases ##j< i-2\, , \,j > i-1## and ##j \in \{i-2,i-1\}## if read from left to right? It is no recursive definition but an explicit formula, so why use induction?
 
Hold on! I forgot that I proved that ##\sigma (x_1,x_2,...,x_n) \sigma^{-1} = (\sigma(x_1),...,\sigma(x_n))##, which makes the problem trivial.
 

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