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A group extension splits exactly when it has a complement or a compatible semidirect-product model, and a kernel retraction forces a direct product

Statement

For a group extension

$1 \to N \overset{i}{\to} E \overset{π}{\to} Q \to 1,$

the following are equivalent:

the extension splits;
$i (N)$ has a complement in $E$ ;
the extension is equivalent to the standard extension $1 \to N \to N ⋊ Q \to Q \to 1$ for some action of $Q$ on $N$ .

If there exists a retraction $r : E \to N$ of the kernel inclusion $i$ , then $\ker r$ is a complement to $i (N)$ , it centralizes $i (N)$ , and therefore $E ≅ N \times \ker r$ .

Facts & Assumptions

Given: The displayed group extension.

[L1]

For a group extension, having a homomorphic section, having a complement to the kernel, and being equivalent to a compatible semidirect-product extension are equivalent (Splitting lemma for groups: a section, a complement, and a semidirect-product decomposition are equivalent).

[L2]

In a group extension, $i (N)$ is normal in $E$ (In a group extension the kernel is normal and the quotient recovers the base).

Proof

technique · iff

1.1givenL1

By [L1], conditions 1, 2, and 3 are equivalent.

1.2givenalgebra

Suppose $r : E \to N$ is a retraction of $i$ . For $x \in i (N) \cap \ker r$ , write $x = i (n)$ . Then $1 = r (x) = r (i (n)) = n$ , so $x = i (1) = 1$ . For any $g \in E$ , the element $i (r (g))^{- 1} g$ lies in $\ker r$ because $r (i (r (g))^{- 1} g) = r (g)^{- 1} r (g) = 1.$ Therefore $g = i (r (g)) (i (r (g))^{- 1} g) \in i (N) \ker r,$ so $E = i (N) \ker r$ and $\ker r$ is a complement to $i (N)$ .

2.1L2step 1.2algebra

Let $k \in \ker r$ and $x = i (n) \in i (N)$ . Because $i (N)$ is normal in $E$ by [L2], the conjugate $k x k^{- 1}$ still lies in $i (N)$ . Applying $r$ gives $r (k x k^{- 1}) = r (k) r (x) r (k)^{- 1} = n .$ Since $r$ restricts to the inverse of $i$ on $i (N)$ , this forces $k x k^{- 1} = i (n) = x$ . Hence $\ker r$ centralizes $i (N)$ , and step 1.2 upgrades the decomposition to a direct product $E ≅ i (N) \times \ker r ≅ N \times \ker r$ .

3.1step 1.1step 1.2step 2.1∎

Step 1.1 is the splitting criterion, while steps 1.2 and 2.1 show that any kernel retraction forces a direct-product splitting.

Depends on

Used by

Extensions with coprime kernel and quotient split Corollary
The pure braid extension splits as a semidirect product Corollary
The zero H² class is equivalent to splitting Corollary
The dihedral group of order eight is a split extension of C₄ by C₂ Example
FALSE: every split group extension is a direct product False statement
A complement determines the conjugation action on the kernel Lemma
If the kernel is complete, the extension splits over its centralizer Proposition
Schur-Zassenhaus existence theorem Theorem

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Group Theory (standard reference, not scraped)