1 Computation

1 2: compute everything

3 4 5: express as disjoint cycles and lcm them.

2 Computation’

compute

3 Sign

1. Change of variable (let \(y_i = h(x_i)\)).

2. Substitution (sub \(x_i = f(i)\)).

4 Odd even

1. compute

2. Because \(\operatorname{sgn}\) is a function

3. because \((12)(12) = \operatorname{id}\)

4. because \(\left\lvert A_n \right\rvert = \left\lvert C \right\rvert\) and by part 2 each must be half as big as \(S_n\).

5 Only 2 cosets implies normal subgroup

Let any \(n\in N\), then for any \(g \in G\), we have two case,

1. \(gn \in N = eN\),
2. \(gn \in C\) where \(C\) is the “other” coset.

We consider the term \(gng^{-1}\), in the first case, \[\begin{align*} ng^{-1} &= ( \underbrace{gn^{-1}}_{\text{in }N} )^{-1} \\ gng^{-1} &= g n' \text{ for some }n' \in N \\ &\in N \end{align*}\] and in the second case, we first claim that \(C = gN = g^{-1} N\), for suppose not, then \(g^{-1}N = N\) which entails that \(eN = g(g^{-1}N) = C\). \[\begin{align*} ng^{-1} &= ( \underbrace{gn^{-1}}_{\text{in }C} )^{-1} \\ &= g^{-1} n' \text{ for some }n' \in N \\ gng^{-1} &= n' \in N \end{align*}\] \(N\) is normal in both cases, as the previous question demonstrated that the only cosets of \(A_n\) are itself and \(C\), we have \(A_n\) is a normal subgroup.

6 Conjugates.

1. Refl, Sym, Trans.

2. Suppose \(g' \sim g\), then unravel definition of \(C_G(g)\) to show that \(g' \in C_G(g)\).

3. Straight from part i

7

By normality, the condition \(C_G(g) \subseteq N\) is satisfied.

8 Cycles

…