Read: Continue with Section 4.3. And if you didn’t last time, consider watching this Socratica video: Cycle Notation of Permutations - Abstract Algebra | Socratica (12:36 min)
Turn in: 4.83, 4.86, 4.91, 4.93
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On 4.83, the notation $\alpha, \beta, \sigma, \gamma$ refers to the permutations drawn in Example 4.78 (but not the $\alpha, \beta$ from the previous paragraph).
Remember that when computing a composition like $\alpha\gamma$, you should work from right to left: $\alpha\gamma(1) = \alpha(\gamma(1))$. But when reading the actual cycle notation, you always read left to right (and then cycle back to the far left). For example, since $\gamma = (1,5)$, $\gamma(1) = 5$, so $\alpha\gamma(1) = \alpha(5)$. Then, using that $\alpha = (1,2,3,4,5)$, we find that $\alpha(5) = 1$, so $\alpha\gamma(1) = 1$. Now let’s check $\alpha\gamma(2)$. Remember that $\gamma = (1,5)$ implicitly means $\gamma = (1,5)(2)(3)(4)$, so $\gamma(2) = 2$. Thus, $\alpha\gamma(2)=\alpha(2)=3$. So far we learned that the cycle notation for $\alpha\gamma$ starts like this $(1)(2,3,\ldots)\cdots$. There are various ways to continue, but typically, we try to continue with the existing cycle by computing $\alpha\gamma(3)$. This time you should find that $\alpha\gamma(3) = 4$, so $\alpha\gamma = (1)(2,3,4,\ldots)\cdots$. Next we compute $\alpha\gamma(4) = 5$, so $(1)(2,3,4,5,\ldots)\cdots$. And then $\alpha\gamma(5)=\alpha(1) = 2$. Since $5$ is sent to the $2$ at the beginning of the cycle, we stop the current cycle that we’re computing to get $(1)(2,3,4,5)\cdots$. There are no more numbers left to compute, so we are done. We also don’t write the cycles of length $1$, so $\alpha\gamma = (2,3,4,5)$ is our answer.
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Make sure to read and memorize 4.85 and 4.89. Both are very important!
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4.93 should be pretty quick using 4.89.