Normalizer of a group and its relation to a subgroup

Quote:

---

Let H be a subgroup of a group G.
a) Show that . Give an example to show that this is not necessarily true if H is not a subgroup of G.

b) Show that if and only if H is abelian.

---

For a) I proved the subgroup property. My problem is the example. H clearly has to be a group of some kind else the statement "H is not a subgroup of N_G(H)" is empty of meaning. So H has to be a group, but not a subgroup of G. I can't image how this could arise. Any hints?

For b) I have proven that if H is abelian then . I am having trouble with "If implies H is abelian. If were abelian it would be easy, but I don't think this is true.

Thanks for any help!
-Dan

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^-1g)H = g^-1(gH) = g^-1Hg, so that:

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

It is quite easy to construct a counter-example: Let G = S₃, 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).

************

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.

Thanks. That did the trick. (One of these days I'm actually going to be able to do stuff like this myself. I'm just not thinking about these proofs the right way.)

-Dan