Hi, I am in difficulty when I kept trying to do these questions.

Could you give me a help to solve these questions?

Thank you so much.

Note: reference below

Q3: Recall that Theorem 3.9.11 (Lagrange's Theorem) states that if H is a subgroup of G, then 003) is equal to 0(H)[G : H]; thus, the order of H divides the order of G. (a) If G 2 S4, then the order of every subgroup of 84 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 justication is ne.) (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 or be a cycle of length 3 in A4, and consider the cosets (TH and 02H in light of Theorem 3.9.11 (Lagrange's Theorem]. In which coset of H is 6'? Q4: Let A = Z3. For any element [a] E U3, we can dene the function f[:;] : Z3 > Z3 by MUM) = [allkl- Let G be the set of all such functions flu]; it is relatively straightforward to show that G is a subgroup of 33,, (the bijections of g). As you might imagine, G is basically just Us, so in this question we will simply conate G and Us. In particular, we can see that g] 0 jib] : ung, so G' ends up being L'the same\" group as U3. (In the language of Section 3.12, G and Us are isomorphic.) For all [k] E Z3, compute the orbit of [11:] under Us, Ug([k]], and the stabilizer of [k], (Ug)[k]. Hint: Using certain information about the orbits and stabilizers, you can save yourself a nontrivial amount of work! You might also think about grouping some elements of EB together. Theorem 3.9.11 (Lagrange's Theorem). Let G be a finite group and let H be a subgroup. Then we have o( G) = o( H ) [G : H]