I would think that the problem in (a) would refer to a a SET H, where we can still form the set:

N(H) = {g in G: gH = Hg}.

It is not hard to show that this is still a subgroup of G:

For g,g' in N(H), we have:

(gg')H = g(g'H) = g(Hg') = (gH)g' = (Hg)g' = H(gg').

Now, suppose g is in N(H), so gH = Hg.

Then H = eH = (g^{-1}g)H = g^{-1}(gH) = g^{-1}Hg, so that:

g^{-1}H = g^{-1}H(gg^{-1}) = (g^{-1}Hg)g^{-1}= Hg^{-1}, so g^{-1}is in N(H).

It is quite easy to construct a counter-example: Let G = S_{3}, with H = {(1 2),(2 3)}. Then:

H(1 2) = {e, (1 3 2)} ≠ (1 2)H = {e, (1 2 3)}

H(2 3) = {(1 2 3), e} ≠ (2 3)H = {(1 3 2), e}

H(1 3) = {(1 3 2), (1 2 3)} = (1 3)H = {(1 2 3), (1 3 2)}

H(1 2 3) = {(2 3), (1 3)} ≠ (1 2 3)H = {(1 3), (1 2)}

H(1 3 2) = {(1 3), (1 2)} ≠ (1 3 2)H = {(2 3), (1 3)}

so we see that N_{G}(H) = {e, (1 3)}, and H is not contained in this subgroup of G.

(the main reason this fails is that an arbitrary subset need not be closed under multiplication, so need not be self-conjugate:

hHh^{-1}might not be H...this can't happen if H is a subgroup since hHh^{-1}= (hH)h^{-1}= Hh^{-1}= H, for a subgroup H).

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OK, for your second question:

Suppose we have H is contained in C_{G}(H). This means that every element of H commutes with every OTHER element of H (since C_{G}(H) is the subset of elements of G that commute with EVERY element of H).

So, we have for any h,h' in H:

hh' = h'h, that is: H is abelian.