Exercise 1.10. Verify that with sigma=(12),tau=(123), the standard representation has a basis alpha=(omega,1,omega^(2)),beta=(1,omega,omega^(2)), with tau alpha=omega alpha,quad tau beta=omega^(2)beta,quad sigma alpha=beta,quad sigma beta=alpha

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Exercise 1.10. Verify that with  sigma=(12),tau=(123), the standard representation has a basis  alpha=(omega,1,omega^(2)),beta=(1,omega,omega^(2)), with  tau alpha=omega alpha,quad tau beta=omega^(2)beta,quad sigma alpha=beta,quad sigma beta=alpha

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Problem Understanding: Understand the problem and the notation. We are given two permutations sigma and tau, and two vectors alpha and beta. We need to verify that these vectors satisfy certain properties when acted upon by the permutations. The permutations are given in cycle notation, and omega is a complex cube root of unity, meaning omega^3 = 1 and omega ≠ 1. The properties to verify are: tau * alpha = omega * alpha, tau * beta = omega^2 * beta, sigma * alpha = beta, sigma * beta = alpha.Calculate tau * alpha: Calculate tau * alpha. The permutation tau = (123) means that the first element goes to the second position, the second goes to the third, and the third goes back to the first. Applying tau to alpha gives us: tau * alpha = (omega^2, omega, 1). Since omega is a cube root of unity, multiplying by omega gives us: omega * alpha = (omega^2, 1, omega). Comparing the two results, we see that tau * alpha = omega * alpha.Calculate tau * beta: Calculate tau * beta. Applying tau to beta gives us: tau * beta = (omega^2, 1, omega). Multiplying beta by omega^2 gives us: omega^2 * beta = (omega^2, 1, omega). Comparing the two results, we see that tau * beta = omega^2 * beta.Calculate sigma * alpha: Calculate sigma * alpha. The permutation sigma = (12) means that the first element swaps with the second, and the third element remains unchanged. Applying sigma to alpha gives us: sigma * alpha = (1, omega, omega^2). This is exactly the vector beta, so sigma * alpha = beta.Calculate sigma * beta: Calculate sigma * beta. Applying sigma to beta gives us: sigma * beta = (omega, 1, omega^2). This is exactly the vector alpha, so sigma * beta = alpha.

Verify that with $\sigma=(1\,2)$ , $\tau=(1\,2\,3)$ , the standard representation has a basis $\alpha=(\omega,1,\omega^{2})$ , $\beta=(1,\omega,\omega^{2})$ , with $\tau \alpha=\omega \alpha$ , $\tau \beta=\omega^{2}\beta$ , $\sigma \alpha=\beta$ , $\sigma \beta=\alpha$

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