In a document, answer the following questions:

In a document, answer the following questions:
Answer 1:
Let Q denote H intersection N and q be an element in it.
To prove: hqh-1 belongs to Q.
Proof: Clearly q belongs to H, thus hqh-1 is in H as H is closed.
Since, N is normal gqg-1 is in N. Since H is a subgroup of G hqh-1 is in N.
Thus, hqh-1 is in Q. Thus Q is normal subgroup of H.
If H is normal in G we will have that H intersection N is normal in G.
Answer 2:
Now, gHg-1 is a subgroup of order 4 as.
gHg-1 = {ghg-1 : h belongs to H}
Clearly the identity is e = geg-1 belongs to gHg-1 .
If x and y belong to gHg-1 , then x = gh1g-1 , y = gh2g-1 for some elements
h1 & h2 in H . Thus, xy-1 = gh1g-1gh2-1g-1 = gh1h2-1g-1 belongs to gHg-1 . So gHg-1 is a subgroup.
Since H is the only subgroup of order 4 and since gHg-1 is a subgroup
of order 4, then gHg-1  and so H is normal by the property that if gHg-1  then H is normal in G .
Answer 3:
Define a homomorphism Φ: R-> C by Φ(Θ)= cis(2piΘ). It is easy to check that Φ is indeed a homomorphism. The map Φ is surjective since every element of C is of the form cis (α) = cis(2piα/2pi)= Φ(α/2pi) for some α in R. The kernel of Φ is the set of Θ in R such that cis(2piΘ) =1.This equation holds true if and only if 2piΘ = 2pik for some k in Z. Dividing by 2pi we see that ker(Φ) =Z, so the first isomorphism theorem tells us that R/Z is isomorphic to the unit circle.
Answer 4:
H={1; (1 2); (1 3); (2 3); (1 2 3); (1 3 2)}
Normaliser of subgroup P generated by (1,2): We know that P is not normal, so its normalizer NS3(P)...