Consider the following permutations in S9: σ=(123456789352146978) and τ=(123456789526713894) Find α by solving σα=τ. Write your answer in cycle notation.

Let σ=(123456789352146978),τ=(123456789526713894) σ−1=(352146978123456789) σα=τ⇒α=δ−1τ ∴α=(123456789765123948) α=(17984)(2635) Let σ=(123456789352146978),τ=(123456789526713894) σ−1=(352146978123456789) σα=τ⇒α=σ−1τ ∴α=(123456789765123948) α=(17984)(2635)

"Consider the following permutations in S9: σ=(123456789352146978) and..." was answered by our experts

College
Algebra
Abstract algebra

uheaeb56e

Answered question

2022-02-14

Consider the following permutations in

S_{9}

σ = (\begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 3 & 5 & 2 & 1 & 4 & 6 & 9 & 7 & 8 \end{matrix})

and

τ = (\begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 5 & 2 & 6 & 7 & 1 & 3 & 8 & 9 & 4 \end{matrix})

Find

α

by solving

σ α = τ

. Write your answer in cycle notation.

Answer & Explanation

dimljenihhml

Beginner2022-02-15Added 15 answers

Let $σ = (\begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 3 & 5 & 2 & 1 & 4 & 6 & 9 & 7 & 8 \end{matrix}), τ = (\begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 5 & 2 & 6 & 7 & 1 & 3 & 8 & 9 & 4 \end{matrix})$
$σ^{- 1} = (\begin{matrix} 3 & 5 & 2 & 1 & 4 & 6 & 9 & 7 & 8 \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \end{matrix})$
$σ α = τ \Rightarrow α = δ^{- 1} τ$
$∴ α = (\begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 7 & 6 & 5 & 1 & 2 & 3 & 9 & 4 & 8 \end{matrix})$
$α = (17984) (2635)$
Let $σ = (\begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 3 & 5 & 2 & 1 & 4 & 6 & 9 & 7 & 8 \end{matrix}), τ = (\begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 5 & 2 & 6 & 7 & 1 & 3 & 8 & 9 & 4 \end{matrix})$
$σ^{- 1} = (\begin{matrix} 3 & 5 & 2 & 1 & 4 & 6 & 9 & 7 & 8 \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \end{matrix})$
$σ α = τ \Rightarrow α = σ^{- 1} τ$
$∴ α = (\begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 7 & 6 & 5 & 1 & 2 & 3 & 9 & 4 & 8 \end{matrix})$
$α = (17984) (2635)$

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