LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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The identified subgroup used to form a central product is central, hence normal

Statement

Let $G$ and $H$ be groups with central subgroups $Z_{1} \leq Z (G)$ and $Z_{2} \leq Z (H)$ and an isomorphism $α : Z_{1} \to Z_{2}$ . The subgroup $N = {(z, α (z)^{- 1}) : z \in Z_{1}}$ of $G \times H$ is central, hence normal, so the quotient $(G \times H) / N$ of The central product $G \circ_{α} H$ of two groups along an isomorphism of central subgroups is defined.

Facts & Assumptions

Given: Groups $G, H$ , central subgroups $Z_{1} \leq Z (G)$ and $Z_{2} \leq Z (H)$ , and an isomorphism $α : Z_{1} \to Z_{2}$ .

[F1]

For groups $G, H$ with central subgroups $Z_{1} \leq Z (G)$ , $Z_{2} \leq Z (H)$ and an isomorphism $α : Z_{1} \to Z_{2}$ , the central product $G \circ_{α} H$ is the quotient of $G \times H$ by $N = {(z, α (z)^{- 1}) : z \in Z_{1}}$ (The central product $G \circ_{α} H$ of two groups along an isomorphism of central subgroups).

[F2]

A subset $H \subseteq G$ is a subgroup when $e \in H$ , $H$ is closed under the operation, and $H$ is closed under inverses (Subgroup).

[F3]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F4]

A subgroup $N \leq G$ is normal in $G$ when $g N g^{- 1} = N$ for every $g \in G$ , where $g N g^{- 1} : = {g n g^{- 1} : n \in N}$ (Normal subgroup: invariance under conjugation).

[L1]

A group homomorphism $f$ satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ (A group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ , and $f (g^{n}) = f (g)^{n}$ for every $n \in Z$ ; for monoid homomorphisms preservation of the identity must be assumed).

[L2]

The external direct product $G \times H : = {(g, h) : g \in G, h \in H}$ carries the componentwise operation $(g, h) (g^{'}, h^{'}) = (g g^{'}, h h^{'})$ (The external direct product $G \times H$ with componentwise multiplication).

[L3]

The componentwise operation makes $G \times H$ a group with identity $(e_{G}, e_{H})$ and $(g, h)^{- 1} = (g^{- 1}, h^{- 1})$ ( $G \times H$ is a group with identity $(e_{G}, e_{H})$ , coordinatewise inverses, and homomorphic coordinate projections).

Proof

technique · direct

1.1

The isomorphism $α$ is in particular a homomorphism, so $α (e) = e$ and $α (z^{- 1}) = α (z)^{- 1}$ ; moreover $Z_{2} \leq Z (H)$ , so any two elements of $Z_{2}$ commute and $α (z_{1} z_{2})^{- 1} = α (z_{1})^{- 1} α (z_{2})^{- 1}$ .

F3L1algebra

2.1

Hence $(e, e) = (e, α (e)^{- 1})$ lies in $N$ ; the product $(z_{1}, α (z_{1})^{- 1}) (z_{2}, α (z_{2})^{- 1}) = (z_{1} z_{2}, α (z_{1})^{- 1} α (z_{2})^{- 1}) = (z_{1} z_{2}, α (z_{1} z_{2})^{- 1})$ lies in $N$ ; and $(z, α (z)^{- 1})^{- 1} = (z^{- 1}, α (z)) = (z^{- 1}, α (z^{- 1})^{- 1})$ lies in $N$ . So $N$ is a subgroup of $G \times H$ .

F1F2L2L3step 1.1

3.1

Every element of $N$ has first coordinate in $Z_{1} \leq Z (G)$ and second coordinate in $Z_{2} \leq Z (H)$ , and the operation is componentwise, so $(z, α (z)^{- 1}) (g, h) = (z g, α (z)^{- 1} h) = (g z, h α (z)^{- 1}) = (g, h) (z, α (z)^{- 1})$ for every $(g, h)$ ; thus $N \leq Z (G \times H)$ .

F1F3L2step 2.1

4.1

For $n \in N$ and $x \in G \times H$ centrality gives $x n x^{- 1} = n x x^{- 1} = n$ , so $x N x^{- 1} = N$ and $N$ is normal; the quotient $(G \times H) / N$ is therefore defined.

F4step 3.1∎

Remarks

Centrality is used twice. It makes the inverse-coordinate rule $z \mapsto (z, α (z)^{- 1})$ multiplicative, so that $N$ is a subgroup, and it then makes that subgroup central and hence normal. For merely isomorphic subgroups the displayed antidiagonal need be neither a subgroup nor a normal subset, so the quotient construction does not apply.

Depends on

Used by

Cited to discharge well-definedness by The central product G∘_α H of two groups along an isomorphism of central subgroups.

Dependency tree · two levels

29 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

M. van Beek, Topics in Finite p-Groups, Definition 2.34 (standard reference, not scraped)
D. A. Craven, The Theory of p-Groups, Theorem 3.6 (standard reference, not scraped)