Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.2 Exercise 4.2.2
Solution: We use the notation $\sigma(k) = \sigma \cdot k$.
(1) $1 \mapsto 1$
(2) We have $$(1\ 2)(1) = (1\ 2) = 2,$$ $$(1\ 2)(2) = (1\ 2)(1\ 2) = 1,$$ $$(1\ 2)(3) = (1\ 2)(2\ 3) = (1\ 2\ 3) = 5,$$ $$(1\ 2)(4) = (1\ 2)(1\ 3) = (1\ 3\ 2) = 6,$$ $$(1\ 2)(5) = (1\ 2)(1\ 2\ 3) = (2\ 3) = 3,$$ and $$(1\ 2)(6) = (1\ 2)(1\ 3\ 2) = (1\ 3) = 4.$$ Thus $(1\ 2) \mapsto (1\ 2)(3\ 5)(4\ 6)$.
(3) We have $$(2\ 3)(1) = (2\ 3) = 3,$$ $$(2\ 3)(2) = (2\ 3)(1\ 2) = (1\ 3\ 2) = 6,$$ $$(2\ 3)(3) = (2\ 3)(2\ 3) = 1,$$ $$(2\ 3)(4) = (2\ 3)(1\ 3) = (1\ 2\ 3) = 5,$$ $$(2\ 3)(5) = (2\ 3)(1\ 2\ 3) = (1\ 3) = 4,$$ and $$(2\ 3)(6) = (2\ 3)(1\ 3\ 2) = (1\ 2) = 2.$$ Thus $(2\ 3) \mapsto (1\ 3)(2\ 6)(4\ 5)$.
(4) We have $$(1\ 3)(1) = (1\ 3) = 4,$$ $$(1\ 3)(2) = (1\ 3)(1\ 2) = (1\ 2\ 3) = 5,$$ $$(1\ 3)(3) = (1\ 3)(2\ 3) = (1\ 3\ 2) = 6,$$ $$(1\ 3)(4) = (1\ 3)(1\ 3) = 1,$$ $$(1\ 3)(5) = (1\ 3)(1\ 2\ 3) = (1\ 2) = 2,$$ and $$(1\ 3)(6) = (1\ 3)(1\ 3\ 2) = (2\ 3) = 3.$$ Thus $(1\ 3) \mapsto (1\ 4)(2\ 5)(3\ 6)$.
(5) We have $$(1\ 2\ 3)(1) = (1\ 2\ 3) = 5,$$ $$(1\ 2\ 3)(2) = (1\ 2\ 3)(1\ 2) = (1\ 3) = 4,$$ $$(1\ 2\ 3)(3) = (1\ 2\ 3)(2\ 3) = (1\ 2) = 2,$$ $$(1\ 2\ 3)(4) = (1\ 2\ 3)(1\ 3) = (2\ 3) = 3,$$ $$(1\ 2\ 3)(5) = (1\ 2\ 3)(1\ 2\ 3) = (1\ 3\ 2) = 6,$$ and $$(1\ 2\ 3)(6) = (1\ 2\ 3)(1\ 3\ 2) = 1.$$ Thus $(1\ 2\ 3) \mapsto (1\ 5\ 6)(2\ 4\ 3)$.
(6) We have $$(1\ 3\ 2)(1) = (1\ 3\ 2) = 6,$$ $$(1\ 3\ 2)(2) = (1\ 3\ 2)(1\ 2) = (2\ 3) = 3,$$ $$(1\ 3\ 2)(3) = (1\ 3\ 2)(2\ 3) = (1\ 3) = 4,$$ $$(1\ 3\ 2)(4) = (1\ 3\ 2)(1\ 3) = (1\ 2) = 2,$$ $$(1\ 3\ 2)(5) = (1\ 3\ 2)(1\ 2\ 3) = 1,$$ and $$(1\ 3\ 2)(6) = (1\ 3\ 2)(1\ 3\ 2) = (1\ 2\ 3) = 5.$$ Thus $(1\ 3\ 2) \mapsto (1\ 6\ 5)(2\ 3\ 4)$.
