Homework 19

Read: Finish Section 4.3, skim 4.4, and read 4.5.

Turn in: 4.95(a), 4.96, 4.97, 4.112, 4.113, 4.115, 4.116, 4.119 (only prove that $A_n$ is a group), 4.120

• Please read Theorem 4.99 and Corollary 4.100—they’re important.
• On 4.112–4.116, you should be able to make use of what you found in 4.97.
• On 4.120, $A_3$ is simply the list of permutations you wrote down in 4.112. Now you have to determine what familiar group it’s isomorphic to.

Remark about even/odd permutations: the point of Theorem 4.110 and Definition 4.111 (which are very important) is that if $\alpha$ is any given permutation and you can find some way to write $\alpha$ as a product of an even (respectively, odd) number of transpositions, then every possible way to write $\alpha$ as a product of transpostions will involve an even (respectively, odd) number. For example, since $(1,2,3)(5,6) = (1,3)(1,2)(5,6)$, we say $(1,2,3)(5,6)$ is an odd permutation because we wrote $(1,2,3)(5,6)$ as a product of three transpositions; notice that $(1,2,3)(5,6)$ can also be written as $(1,2,3)(5,6) = (5,6)(1,2)(5,6)(2,3)(5,6)$ (and lots of other ways, but always using an odd number of transpositions).

Remark about inverses in \(S_n\): it’s useful to know how to quickly compute the inverse of a permutation in cycle notation. First, just think about a single cycle like $\alpha = (1,4,5,7)$. Think about where $\alpha$ sends each number; now, where should $\alpha^{-1}$ send each number? Try to convince yourself that $\alpha^{-1}$ can be written by just reversing the cycle notation for $\alpha$ to get $\alpha^{-1}=(7,5,4,1)$. And if you want, you can rewrite $(7,5,4,1)$ as $(1,7,5,4)$.

Now, more generally, if $\beta$ is written as a product of cycles (disjoint or not), you can use Theorem 2.44 (“socks and shoes”) to see that $\beta^{-1}$ can be written by reversing the order of the cycles forming $\beta$ and then reversing the order of the numbers in each of the cycles (like we did with $\alpha$ above). For example, if $\beta = (1,4,5,7)(2,9)(3,6,8)$, then $\beta^{-1} = (3,6,8)^{-1}(2,9)^{-1}(1,4,5,7)^{-1} = (8,6,3)(9,2)(7,5,4,1)$.

Extra practice: 4.98, 4.99, 4.106, 4.107, 4.121