Homework 19

Read: Continue with Section 4.3. Consider watching this Socratica video: Cycle Notation of Permutations - Abstract Algebra | Socratica (12:36 min)

Turn in: 4.77 or 4.78, 4.79(a,b,e,f,g,k,l), 4.83, 4.86, 4.87, 4.89, 4.91(a)

• On 4.79, the notation $\alpha, \beta, \sigma, \gamma$ refers to the permutations drawn at the bottom of page 67 (but not the $\alpha, \beta$ from the previous paragraph).

  Also, remember that when computing a composition like $\alpha\gamma$, you should work from right to left: $\alpha\gamma(1) = \alpha(\gamma(1))$. However, when reading cycle notation, you always read left to right (and then cycle back to the far left)—a little confusing, right?! 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.

• Problem 4.87 should be really short if you make use of Theorem 4.85.

Extra practice: 4.81, 4.82, 4.84, 4.85

• On 4.81, you want to show that $\alpha\beta=\beta\alpha$ for arbitrary disjoint cycles $\alpha$ and $\beta$ in $S_n$. To do this, you should think of the permutations as functions. Thus, $\alpha\beta=\beta\alpha$ if and only if $\alpha\beta(i)=\beta\alpha(i)$ for every $1\le i\le n$. So want to show $\alpha\beta(i)=\beta\alpha(i)$ for every $1\le i\le n$, and now you want to make use of the fact that $\alpha$ and $\beta$ are disjoint.