The Normalizer and Normal Closure Subgroups

In this section we are going to ask ourselves: Given a subgroup H of G, can we find a subgroup N of G for which H lies inside of N as a normal subgroup?

Definition 7.12.1. Let G be a group, let H ⊆ G be a subgroup, and define the normalizer of H to be

NG(H) = {g ∈ G | Hg = H} = {g ∈ G | gH−1g = H}

Proposition 7.12.2. NG(H) is a subgroup of G.

Proof.

  1. (NG(H) nonempty): Trivially eHe−1 = H, so e ∈ NG(H).

  2. (Closure): Given x,y ∈ NG(H)

    (xy) H (xy)−1 = xy H y−1x−1 = x−1Hx = H

    So xy ∈ NG(H). (We used the fact that (xy)−1 = y−1x−1, proposition Corollary 7.1.4).

  3. (Existence of the inverse): Given an x ∈ NG(H):

    xHx−1 = H

    By multiplying to ther right for x

    xH = Hx   multiplying to the left by x−1   H = x−1 Hx

    thus x ∈ NG(H).

Thus NG(H) is a subgroup of G as desired.   □.

Proposition 7.12.3. H ⊲̲ NG(H). We have H ⊲̲ G only if NG(H) = G.

Proof. First, we must check to see that H is a normal subgroup of NG(H). But this is obvious, since g−1 Hg = H for all g ∈ NG(H)

NG(H) = G just outright means that Hx = H, ∀x ∈ G: That's exactly normality of H.  □

Lemma 7.12.4. Let N1, N2 be normal subgroups of the group G. Then:

  1. N1 ∩ N2 is a normal subgroup of G.

  2. N1N2 is a normal subgroup of G.

  3. If H is any subgroup of G then N1 ∩ H is a normal subgroup of H and N1H = HN1 and hence N1H is a subgroup of G.

Proof. (1) Let n ∈ N1 ∩ N2 and g ∈ G. Then g−1ng ∈ N1, since N, is normal. Similarly g−1ng ∈ N2 since N2 is normal. Hence g−1ng ∈ N1 ∩ N2. It follows that g−1(N1 ∩ N2)g ⊂ N1 ∩ N2 and therefore N1 ∩ N2 is normal.
(2) Let n ∈ N1, n2 ∈ N2. Since N1, N2 are both normal N1N2 = N2N1, as sets and N1N2 forms a subgroup of G by Theorem 7.9.7 Let g ∈ G and n1n2 ∈ N1N2. Then

g−1(n1n2)g = (g−1n1g)(g−1n2g) ∈ N1N2

since g−1n1g ∈ N1, and g−1n2g ∈ N1 ∈ N2. Therefore N1N2 is normal in G.
(3) Let h ∈ H and n ∈ N1 ∩ H. Then h−1nh ∈ N1, because N1 ⊲ G and h−1nh ∈ H because h,n both do and therefore h−1nh ∈ N1 ∩ H. Hence N1 ∩ H is a normal subgroup of H. But since N1 ⊲ G it follows that N1H = HN1. Again by Theorem 7.9.7, this implies that NH is a subgroup of G.  □.

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