Q3: Recall that Theorem 3.9.11 (Lagrange's Theorem) states that if H is a subgroup of G, then o(G) is equal to o(H)[G : H]; thus, the order of H divides the order of G. (a) If G = S4, then the order...

Extracted text: Q3: Recall that Theorem 3.9.11 (Lagrange's Theorem) states that if H is a subgroup of G, then o(G) is equal to o(H)[G : H]; thus, the order of H divides the order of G. (a) If G = S4, then the order of every subgroup of S4 must divide 4! = 24. For every distinct divisor of 24, is there a subgroup of S4 with that order? (You don't have to go through a big formal proof that each of your claimed subgroups are actually subgroups; very brief justification is fine.) (b) Prove that the group A4 does not have a subgroup of order 6. Hint: Suppose, for contradiction, that H is a subgroup of A4 with order 6. Let o be a cycle of length 3 in A4, and consider the cosets oH and o²H in light of Theorem 3.9.11 (Lagrange's Theorem). In which coset of H is o?