Question

Let σ = (a1 a2 ...ak) ∈ Sn be a cycle let τ ∈ Sn.
i) Check that τ σ τ^−1 = (b1 b2 ···bk), where τ (ai) = bi.
ii) Use the above result to compute τ σ τ^−1 where σ and τ are as in part b).

Answer 1

Answer on Question #44737 – Math – Abstract Algebra:

Let $\sigma = (a_{1} a_{2} \ldots a_{k}) \in S_{n}$ be a cycle. Let $\tau \in S_{n}$. Check that $\tau \sigma \tau^{-1} = (b_{1} b_{2} \cdots b_{k})$, where $\tau(a_{i}) = b_{i}$.

Solution.

Denote $\alpha = \tau \sigma \tau^{-1}$. We need to prove that:

$$\forall i = 1, \dots , k - 1: \alpha (b _ {i}) = b _ {i + 1};$$ $$\alpha (b _ {k}) = b _ {1};$$ $$\forall x \in \{1, \dots , n \} \backslash \{b _ {1}, \dots , b _ {k} \}: \alpha (x) = x;$$

So:

$$\begin{array}{l} 1 \leq i \leq k - 1 \Rightarrow \alpha (b _ {i}) = (\tau \sigma \tau^ {- 1}) (b _ {i}) = \tau (\sigma (\tau^ {- 1} (b _ {i}))) = \tau (\sigma (a _ {i})) = \tau (a _ {i + 1}) = b _ {i + 1}; \\ \alpha (b _ {k}) = (\tau \sigma \tau^ {- 1}) (b _ {k}) = \tau (\sigma (\tau^ {- 1} (b _ {k}))) = \tau (\sigma (a _ {k})) = \tau (a _ {1}) = b _ {1}; \\ x \notin \{b _ {1}, \dots , b _ {k} \}, x = \tau (y) \Rightarrow y \notin \{a _ {1}, \dots a _ {k} \} \Rightarrow \sigma (y) = y \Rightarrow \\ \Rightarrow \alpha (x) = (\tau \sigma \tau^ {- 1}) (x) = \tau (\sigma (\tau^ {- 1} (x))) = \tau (\sigma (y)) = \tau (y) = x. \end{array}$$

www.AssignmentExpert.com