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Normalizer quotients, kernels and conditional surjectivity

Statement

Let $H$ be a subgroup of a group $G$ . Its normalizer is $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ . Then $H$ is normal in $N_{G} (H)$ , so the quotient group $N_{G} (H) / H$ exists.

Let $p : G \to Q$ be a group homomorphism and put $K = p (H)$ . The rule $\overset{ˉ}{p} (g H) = p (g) K$ defines a homomorphism from $N_{G} (H) / H$ to $N_{Q} (K) / K$ . Its kernel is ${a H : a \in \ker p \cap N_{G} (H)}$ , and its image is $p (N_{G} (H)) / K$ . In particular, $\overset{ˉ}{p}$ is surjective if and only if $p (N_{G} (H)) = N_{Q} (K)$ .

More generally, if $π : E \to B$ is a surjective group homomorphism with kernel $M$ and $C \leq E$ , then $π ∣_{C} : C \to B$ is an isomorphism exactly when $E = M C$ and $M \cap C = {1}$ . Under these conditions $E$ is isomorphic to $M ⋊ B$ by a map compatible with the inclusion of $M$ and the quotient map, with action induced by conjugation through $(π ∣_{C})^{- 1}$ .

Statement references

The normalizer N_G(H)={g∈ G:gHg⁻¹=H} of a subgroup

C_G(x) and N_G(H) are subgroups of G

The quotient group G/N and coset product (gN)(hN)=ghN

First isomorphism theorem for groups: G/ker f congimf

Splitting lemma for groups: a section, a complement, and a semidirect-product decomposition are equivalent

Monoid homomorphism and group homomorphism

A group homomorphism automatically satisfies f(e) = e' and f(g⁻¹) = f(g)⁻¹, and f(gⁿ) = f(g)ⁿ for every n ∈ ℤ; for monoid homomorphisms preservation of the identity must be assumed

The kernel and image of a group homomorphism

Normal subgroup: invariance under conjugation

The image of a group homomorphism is a subgroup and its kernel is a normal subgroup

Group and abelian group

Subgroup

Group isomorphisms, automorphisms and the set Aut(G)

The external semidirect product N⋊_α H

Group extensions, sections, complements, and split extensions

The inverse of a bijective group homomorphism is a group homomorphism

Left and right cosets gH and Hg of a subgroup

Injection, surjection, bijection

An action of a group H on a group N by automorphisms

Facts & Assumptions

[F1]

The normalizer consists of precisely the elements that conjugate the subgroup onto itself; it is a subgroup. (The normalizer N_G(H)={g∈ G:gHg⁻¹=H} of a subgroup, C_G(x) and N_G(H) are subgroups of G)

[F2]

A left coset is $g H = {g h : h \in H}$ . For a normal subgroup, quotient elements are cosets and their product is $(g H) (h H) = g h H$ . (Left and right cosets gH and Hg of a subgroup, The quotient group G/N and coset product (gN)(hN)=ghN)

[F3]

A group homomorphism induces an isomorphism from its quotient by its kernel onto its image. (First isomorphism theorem for groups: G/ker f congimf)

[F4]

For a surjective group homomorphism, a complement to its kernel yields a compatible semidirect-product decomposition, with the quotient acting by conjugation through the corresponding section. A complement $C$ satisfies $E = M C$ and $M \cap C = {1}$ . (Splitting lemma for groups: a section, a complement, and a semidirect-product decomposition are equivalent)

[F5]

A group homomorphism preserves products, the identity and inverses. Its restriction to a subgroup is a homomorphism because the same product identity holds on the subgroup. (Monoid homomorphism and group homomorphism, A group homomorphism automatically satisfies f(e) = e' and f(g⁻¹) = f(g)⁻¹, and f(gⁿ) = f(g)ⁿ for every n ∈ ℤ; for monoid homomorphisms preservation of the identity must be assumed)

[F6]

For a homomorphism $f : X \to Y$ , its kernel is ${x \in X : f (x) = 1_{Y}}$ and its image is ${f (x) : x \in X}$ ; it is surjective precisely when its image is $Y$ . (The kernel and image of a group homomorphism, Injection, surjection, bijection)

[F7]

A subgroup $L \leq X$ is normal in $X$ precisely when $x L x^{- 1} = L$ for every $x \in X$ . (Normal subgroup: invariance under conjugation)

[F8]

The image of a group homomorphism is a subgroup of its codomain, and its kernel is normal in its domain. (The image of a group homomorphism is a subgroup and its kernel is a normal subgroup)

[F9]

Groups obey associativity, identity and inverse laws. A subgroup contains the identity and is closed under products and inverses, with the inherited group structure. (Group and abelian group, Subgroup)

[F10]

A map is injective if equal images imply equal inputs, surjective if every target element has a preimage, and bijective if both conditions hold. An isomorphism is a bijective group homomorphism, and its inverse is a group homomorphism. (Injection, surjection, bijection, Group isomorphisms, automorphisms and the set Aut(G), The inverse of a bijective group homomorphism is a group homomorphism)

[F11]

A short exact sequence consists of an injective homomorphism followed by a surjective homomorphism whose kernel is the image of the first. A section $s : B \to E$ is a homomorphism with $π \circ s = {id}_{B}$ ; a complement $C$ satisfies $E = (\ker π) C$ and $(\ker π) \cap C = {1}$ . (Group extensions, sections, complements, and split extensions, Injection, surjection, bijection)

[F12]

An action by automorphisms is a homomorphism $α : B \to Aut (M)$ . For such an action, the external semidirect product $M ⋊_{α} B$ has underlying set $M \times B$ and product $(m, b) (m^{'}, b^{'}) = (m α_{b} (m^{'}), b b^{'})$ . (An action of a group H on a group N by automorphisms, The external semidirect product N⋊_α H)

Proof

Given: Groups $G, Q$ , a subgroup $H \leq G$ , and a homomorphism $p : G \to Q$ , with $K = p (H)$ . Also given are a surjective homomorphism $π : E \to B$ , its kernel $M$ , and a subgroup $C \leq E$ .

Goal: Establish the quotient map, its kernel and image, and the stated splitting criterion.

1.1givenF1F2F5F7F8F9

By [F5], $p ∣_{H} : H \to Q$ is a homomorphism, so [F8] gives $K = p (H) \leq Q$ . Every $h \in H$ conjugates $H$ onto itself, so $H \leq N_{G} (H)$ . By [F1], every element of $N_{G} (H)$ conjugates $H$ onto itself; hence $H$ is normal there by [F7]. The same reasoning applies to the subgroup $K \leq Q$ , so both quotient groups exist by [F2]. For $g \in N_{G} (H)$ , preservation of products and inverses in [F5] gives $p (g) K p (g)^{- 1} = p (g H g^{- 1}) = K$ , and therefore $p (g) \in N_{Q} (K)$ .

1.2givenF3F5F6F9F10

The restriction $π ∣_{C}$ is a homomorphism by [F5]; its kernel is $M \cap C$ , and its image is $π (C)$ by [F6]. By [F3], the restriction is an isomorphism onto $B$ exactly when $M \cap C = {1}$ and $π (C) = B$ . The latter equality is equivalent to $E = M C$ : if $π (C) = B$ , then for each $e \in E$ choose $c \in C$ with $π (c) = π (e)$ ; [F5] gives $π (e c^{- 1}) = 1_{B}$ , so $e c^{- 1} \in M$ and $e \in M C$ . Conversely, applying $π$ to $E = M C$ gives $B = π (E) = π (C)$ because $π (M) = {1_{B}}$ .

2.1step 1.1F2F5F6F7F8given

Define $\overset{ˉ}{p} (g H) = p (g) K$ . If $g H = g^{'} H$ , then $g^{'} = g h$ for some $h \in H$ , and $p (g^{'}) K = p (g) p (h) K = p (g) K$ by [F5]. Thus the rule is well defined; the coset product in [F2] and preservation of products show that it is a homomorphism. By [F5] and [F8] applied to $p ∣_{N_{G} (H)}$ , the image $p (N_{G} (H))$ is a subgroup of $N_{Q} (K)$ . Since $K = p (H) \leq p (N_{G} (H))$ and $K$ is normal in $N_{Q} (K)$ , it is normal in this image subgroup by [F7]. The image of $\overset{ˉ}{p}$ is therefore the quotient group $p (N_{G} (H)) / K$ .

3.1step 1.1step 2.1givenF2F5F6

If $a \in \ker p \cap N_{G} (H)$ , then $a H \in \ker \overset{ˉ}{p}$ by [F6]. Conversely, if $g H \in \ker \overset{ˉ}{p}$ , then $p (g) \in K = p (H)$ ; choose $h \in H$ with $p (h) = p (g)$ . By [F5], $a = g h^{- 1}$ has $p (a) = 1_{Q}$ and lies in $\ker p \cap N_{G} (H)$ , with $a H = g H$ . This proves the kernel formula. Since $K \leq p (N_{G} (H)) \leq N_{Q} (K)$ , equality of the two quotient groups $p (N_{G} (H)) / K$ and $N_{Q} (K) / K$ is equivalent to equality of their numerator subgroups: if their coset sets agree, every $n \in N_{Q} (K)$ has $n K = x K$ for some $x \in p (N_{G} (H))$ , and then $n \in x K \subseteq p (N_{G} (H))$ . The image calculation and [F6] therefore give the stated equivalence for surjectivity.

4.1step 1.2F4F5F8F9F10F11F12given∎

Under the equivalent conditions of step 1.2, $C$ is a complement to $M$ in the sense of [F11]. By [F8], $M = \ker π$ is normal in $E$ . The inclusion $M ↪ E$ is an injective homomorphism with image $\ker π$ , and $π$ is surjective, so these maps form a short exact sequence by [F11]. By [F10], $(π ∣_{C})^{- 1} : B \to C$ is a homomorphism; composing it with the subgroup inclusion gives a homomorphism $s : B \to E$ by [F5]. Since $π \circ s = {id}_{B}$ , it is a section by [F11]. Applying [F4] gives the compatible isomorphism $M ⋊ B ≅ E$ , using the external product of [F12] with action $α_{b} (m) = s (b) m s (b)^{- 1}$ . This completes the proof.

Remarks

Surjectivity of the original map $p$ alone does not assert the normalizer-lifting equality. Its verification in any particular Artin–Coxeter application requires the hypotheses of that application.

Depends on

Used by · 4

Dependency tree · 2 levels

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