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

Re: Normalizer of a group and its relation to a subgroup

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).

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

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.

Re: Normalizer of a group and its relation to a subgroup

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