Solutions to Problem Set 13

Math 345/504

3-2-2018

1. Do the following permutation computations, performing any multiplications from right to left. Express your answers in disjoint cycle form.

(a) $(5\ 1\ 4\ 3)^2$ .

(b) $(1\ 2\ 4)^{-1}(6\ 2\ 4\ 1)$ .

(c) $(2\ 4\ 8\ 5)^{103}$ .

(d) $\left((2\ 3\ 6)(1\ 3\
   4)\right)^2$ .

(a) $(5\ 1\ 4\ 3)^2 = (5\ 4)(1\
   3)$ .

(b) $(1\ 2\ 4)^{-1}(6\ 2\ 4\ 1) =
   (4\ 2\ 1)(6\ 2\ 4\ 1) = (1\ 6)$ .

(c) Since $(2\ 4\ 8\ 5)^4 = \id$ , $(2\ 4\ 8\ 5)^{103} = (2\ 4\ 8\ 5)^3 = (2\ 5\ 8\ 4)$ .

(d) $(2\ 3\ 6)(1\ 3\ 4) = (1\ 6\
   2\ 3\ 4)$ , so $\left((2\ 3\ 6)(1\ 3\ 4)\right)^2 =
   (1\ 2\ 4\ 6\ 3)$ .


2. Find the order of the following permutations. Multiply permutations from right to left.

(a) $(5\ 1\ 3\ 7)$ in $S_7$ .

(b) $(1\ 6) (5\ 2\ 3\ 4)$ in $S_6$ .

(c) $(1\ 3\ 2\ 5)(1\ 4\ 2\ 3)$ in $S_5$ .

(a) $(5\ 1\ 3\ 7)$ is a 4-cycle, so its order is 4.

(b) The cycles are disjoint, so the order of their product is the least common multiple of their orders. The order is $[2, 4] = 4$ .

(c) The cycles are not disjoint, so I multiply first:

$$\matrix{ 1 & 2 & 3 & 4 & 5 & \cr & & & & & (1\ 4\ 2\ 3) \cr 4 & 3 & 1 & 2 & 5 & \cr & & & & & (1\ 3\ 2\ 5) \cr 4 & 2 & 3 & 5 & 1 & \cr}$$

The product is $(1\ 4\ 5)$ , which has order 3.


[MATH 504]

3. Give a specific element of $S_6$ which:

(a) Is a product of two disjoint cycles and has order 3.

(b) Is a product of two disjoint cycles and has order 4.

(c) Has order 2, but does not leave any of $\{1, 2, 3, 4, 5, 6\}$ fixed.

(a) $(1\ 2\ 3)$ has order 3 and $(4\ 5\ 6)$ has order 3. Since the cycles are disjoint, the order of the product is $[3, 3] = 3$ . Therefore, $(1\ 2\ 3)(4\ 5\ 6)$ has order 3.

(b) $(1\ 2)$ has order 2 and $(3\ 4\ 5\ 6)$ has order 4. Since the cycles are disjoint, the order of the product is $[2, 4] = 4$ . Therefore, $(1\ 2)(3\ 4\ 5\ 6)$ has order 4.

(c) $(1\ 2)$ , $(3\ 4)$ , and $(5\ 6)$ all have order 2. Since the cycles are disjoint, the order of the product is $[2, 2,
   2] = 2$ . Therefore, $(1\ 2)(3\ 4)(5\ 6)$ has order 2 and does not leave any of $\{1, 2, 3, 4, 5, 6\}$ fixed.


I am not afraid of storms, for I am learning to sail my own ship. - Louisa May Alcott


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