Compute the inverse of \((465312)\).
Solution: Since \((465312)=(42)(41)(43)(45)(46)\), we have \((465312)^{-1}=(46)(45)(43)(41)(42)=(421356)\). #

Let \(G\) be a group and define a map \(\lambda_g:G\to G\) by \(\lambda_g(a)=ga\). Prove that \(\lambda_g\) is a permutation of \(G\).
Proof: We just need to show that \(\lambda_g\) is bijective.

• \(\lambda_g\) is injective.
  Suppose that \(\lambda_g(a)=\lambda_g(b)\), then \(ga=gb\Rightarrow g^{-1}ga=g^{-1}gb\Rightarrow a=b\). Thus \(\lambda_g\) is injective.
• \(\lambda_g\) is surjective.
  For any \(\alpha\in G\), then \(g^{-1}\alpha\in G\) since \(G\) is a group. Note that \(\lambda_g(g^{-1}\alpha)=gg^{-1}\alpha=\alpha\). Thus \(\lambda_g\) is surjective.

Therefore, \(\lambda_g\) is bijective. #