Question

Please use two methods to prove that the set of odd permutations of Sn is not a subgroup of Sn.

Answer #1

Do the odd permutations of Sn form a subgroup of
Sn? If so, provide a proof. If not, provide a
counterexample with justification

Let n ≥ 2. Show that exactly half of the permutations in Sn are
even , by finding a bijection from the set of all even permutations
in Sn to the set of all odd permutations in Sn.

Suppose that the set An has all the even permutations
in n-permutations. Prove that this set is the same as [a set
consisting of cyclic permutations of length 3 and their
products].

Show that if G is a subgroup of Sn and |G| is odd, then G is a
subgroup of An, given n≥2

Prove thatAnis a normal subgroup of Sn. Prove both that it is a
subgroup AND that it is normal

Prove that the set of odd numbers is infinite.

If kr<=n, where 1<r<=n. Prove that the number of
permutations α ϵ Sn, where α is a product of k disjoint r-cycles is
(1/k!)(1/r^k)[n(n-1)(n-2)...(n-kr+1)]

Prove that if an integer is odd, then its square is also odd.
Use the result to establish that if the square of an integer is
known to be even, the integer must be even

a. Prove for all σ, τ ∈ Sn that στσ−1 τ −1 ∈ An.
b. Let p and q be distinct odd primes. Prove that
Zxpq is not a cyclic group.

Let n be an integer greater than 2. Prove that every subgroup of
Dn with odd order is cyclic.

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