# Normalizer of a group and its relation to a subgroup

• Dec 18th 2013, 09:30 AM
topsquark
Normalizer of a group and its relation to a subgroup
Quote:

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

b) Show that $H \leq C_G(H)$ 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 $H \leq C_G(H)$. I am having trouble with "If $H \leq C_G(H)$ implies H is abelian. If $C_G(H)$ were abelian it would be easy, but I don't think this is true.

Thanks for any help!
-Dan
• Dec 18th 2013, 12:40 PM
Deveno
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-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 NG(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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