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The twisted product of a normalized cocycle is a central extension

Statement

Let $Q$ be a group and let $α : Q \times Q \to C^{\times}$ be a normalized two-cocycle on $Q$ with the trivial action on $C^{\times}$ , in the multiplicative convention $α (1, q) = α (q, 1) = 1$ and $α (q, r) α (q r, s) = α (r, s) α (q, r s)$ . Then $E_{α} : = Q \times C^{\times}, (q, z) (r, w) : = (q r, α (q, r) z w)$ is a group with identity $(1, 1)$ and $(q, z)^{- 1} = (q^{- 1}, α (q, q^{- 1})^{- 1} z^{- 1})$ . The second factor $C^{\times} ≅ {1} \times C^{\times}$ is a central subgroup of $E_{α}$ , the projection $E_{α} \to Q$ is a surjective homomorphism with kernel ${1} \times C^{\times}$ , and $E_{α} / ({1} \times C^{\times}) ≅ Q$ . In particular $E_{α}$ need not be finite.

Facts & Assumptions

Given: A group $Q$ , a normalized two-cocycle $α : Q \times Q \to C^{\times}$ in the multiplicative convention of Normalized two-cocycle and two-coboundary, and the set $E_{α} = Q \times C^{\times}$ with the displayed product.

[F1]

A normalized projective representation with factor set $α$ satisfies $α (1, q) = α (q, 1) = 1$ and $α (q, r) α (q r, s) = α (r, s) α (q, r s)$ for all $q, r, s$ . (The factor set satisfies the two-cocycle equation).

[F2]

Read multiplicatively with trivial action, a normalized two-cocycle on $Q$ is exactly a function $α : Q \times Q \to C^{\times}$ with $α (1, q) = α (q, 1) = 1$ and $α (q, r) α (q r, s) = α (r, s) α (q, r s)$ . (Normalized two-cocycle and two-coboundary).

[F3]

A group is a set with an associative binary operation, a two-sided identity, and two-sided inverses. (Group and abelian group).

[F4]

For a homomorphism $f : G \to H$ , the rule $g \ker f \mapsto f (g)$ is an isomorphism $G / \ker f \to im f$ . (First isomorphism theorem for groups: $G / \ker f ≅ im f$ ).

[F5]

The center $Z (G) = {z \in G : z g = g z for every g \in G}$ consists of the elements commuting with every element of $G$ . (The center $Z (G)$ of a group).

[F6]

The kernel of a group homomorphism is the set of elements mapped to the identity. (The kernel and image of a group homomorphism).

Proof

technique · direct

1.1

The product is associative: for $q, r, s \in Q$ and $z, w, x \in C^{\times}$ , the two bracketing orders of $(q, z) (r, w) (s, x)$ give $(q r s, α (q, r) α (q r, s) z w x)$ and $(q r s, α (r, s) α (q, r s) z w x)$ , and these scalars are equal by the cocycle identity of [F1], [F2]; multiplication in each coordinate is associative as well, so the two results coincide.

F1F2algebra

1.2

The element $(1, 1)$ is a two-sided identity: $(1, 1) (q, z) = (q, α (1, q) z) = (q, z)$ and $(q, z) (1, 1) = (q, α (q, 1) z) = (q, z)$ for all $q, z$ , by the normalization in [F1], [F2].

F1F2algebra

2.1

The element $(q^{- 1}, α (q, q^{- 1})^{- 1} z^{- 1})$ is a two-sided inverse of $(q, z)$ . On the right, $(q, z) (q^{- 1}, α (q, q^{- 1})^{- 1} z^{- 1}) = (1, α (q, q^{- 1}) α (q, q^{- 1})^{- 1} z^{- 1} z) = (1, 1)$ ; on the left, $(q^{- 1}, α (q, q^{- 1})^{- 1} z^{- 1}) (q, z) = (1, α (q^{- 1}, q) α (q, q^{- 1})^{- 1} z^{- 1} z)$ , and the scalar is $1$ because the cocycle identity at $(q, q^{- 1}, q)$ reads $α (q, q^{- 1}) α (1, q) = α (q^{- 1}, q) α (q, 1)$ , that is $α (q, q^{- 1}) = α (q^{- 1}, q)$ by the normalization of [F1], [F2].

F1F2step 1.2algebra

3.1

Steps 1.1, 1.2 and 2.1 exhibit an associative product on $E_{α}$ with a two-sided identity and two-sided inverses, so [F3] makes $E_{α}$ a group.

F3step 1.1step 1.2step 2.1

4.1

The projection $π : E_{α} \to Q$ , $π (q, z) = q$ , is a homomorphism: $π ((q, z) (r, w)) = π (q r, α (q, r) z w) = q r = π (q, z) π (r, w)$ ; it is surjective because $(q, 1) \mapsto q$ , and its kernel is ${1} \times C^{\times}$ by the normalization, a subgroup isomorphic to $C^{\times}$ . That kernel is central: $(1, z^{'}) (q, z) = (q, α (1, q) z^{'} z) = (q, z^{'} z) = (q, α (q, 1) z z^{'}) = (q, z) (1, z^{'})$ for all $q, z, z^{'}$ , so it lies in $Z (E_{α})$ in the sense of [F5]. By [F4] applied to $π$ , the quotient of $E_{α}$ by this kernel, which is normal since centrality gives $e k e^{- 1} = k$ for every $e \in E_{α}$ and every kernel element $k$ , is isomorphic to the image $Q$ . Finally $E_{α}$ is infinite whenever $Q$ is nonempty, since ${q} \times C^{\times}$ is an infinite subset for any $q \in Q$ .

F1F2F4F5F6step 3.1algebra

5.1

Collecting steps 3.1 and 4.1: $E_{α}$ is a group with identity $(1, 1)$ and inverses $(q, z)^{- 1} = (q^{- 1}, α (q, q^{- 1})^{- 1} z^{- 1})$ , whose central subgroup ${1} \times C^{\times} ≅ C^{\times}$ has quotient $E_{α} / ({1} \times C^{\times}) ≅ Q$ , and which is infinite when $Q \neq \emptyset$ ; this is the central extension of $Q$ by $C^{\times}$ determined by $α$ .

step 3.1step 4.1∎

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Sources

Britta Späth, Reduction theorems for some global-local conjectures — Proposition 1.11, printed p. 4 (standard reference, not scraped)
Tammo tom Dieck, Representation Theory — §4.2, printed pp. 54–57 (standard reference, not scraped)