Groups - normaliser problems

Let G be a group and let H be a subgroup of G. The normalizer of H in G, written NG(H) (with G as a subscript), is defined by NG(H) = {g ∈ G| gHg^−1 = H}.
(Remember that gHg^−1 = H is an equality of sets, and does not imply that ghg−1 = h for all h ∈ H.)

(a) Show that NG(H) is a subgroup of G.

It is easy to see that H is a normal subgroup of NG(H); you need not prove this.
Notice that H is a normal subgroup of G if and only if NG(H) = G.

(b) In this part, let G = S4 and H = {1, (1 2 4), (1 4 2)}. List the elements of NG(H), explaining your reasoning.

(c) Returning to general G and H, let X be the set of (left) cosets in G of the subgroup NG(H). Let Y be the set of all subgroups of G which are conjugate to H. (We say that a subgroup H' is conjugate to H if H' = gHg−1 for some g ∈ G; recall that gHg−1 is a subgroup of G for any g ∈ G.) Show that
there is a bijection between X and Y.

(a) Check closure under multiplication in NG(H) and inverses.

Let g, g' ∈ NG(H). So, we have gHg⁻¹ = H and g'H(g')⁻¹ = H.

Therefore,
(gg') H (gg')⁻¹
= (gg') H ((g')⁻¹ g⁻¹)
= g (g' H (g')⁻¹) g⁻¹
= gHg⁻¹, since g' ∈ NG(H)
= H, since g ∈ NG(H)

So, gg' ∈ NG(H).
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Next, note that
g⁻¹ H (g⁻¹)⁻¹
= g⁻¹ H g
= g⁻¹ (gHg⁻¹) g, since g ∈ NG(H)
= (g⁻¹g) H (g⁻¹g)
= eHe
= H.

Hence, g⁻¹ ∈ NG(H).

Thus, NG(H) is a subgroup of G.
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(b) This is a straight computation. You may want to take advantage of the closure under multiplication and inverses to do this more efficiently. I'll get you started.

In this case, NG(H) = {g ∈ S4| gHg⁻¹ = H}.

For (1):
(1) * {(1), (1 2 4), (1 4 2)} * (1)⁻¹ = {(1), (1 2 4), (1 4 2)}.
Since (1) H (1)⁻¹ = H, we have (1) ∈ NG(H).

For (12):
(12) * {(1), (1 2 4), (1 4 2)} * (12)⁻¹ = {(1), (1 4 2), (1 2 4)}.