PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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The two canonical maps into a central product are injective homomorphisms whose images commute, generate it, and meet in the identified centre

Statement

Let $G$ and $H$ be groups with central subgroups $Z_{1} \leq Z (G)$ and $Z_{2} \leq Z (H)$ and an isomorphism $α : Z_{1} \to Z_{2}$ , and let $π : G \times H \to G \circ_{α} H$ be the quotient map. The canonical maps $G \to G \circ_{α} H$ and $H \to G \circ_{α} H$ are injective homomorphisms whose images commute elementwise, generate $G \circ_{α} H$ , and meet in the image of $Z_{1}$ . They are $g \mapsto \overset{ˉ}{g} = π (g, e)$ and $h \mapsto \overset{ˉ}{h} = π (e, h)$ , and the intersection of their images is ${\overset{ˉ}{z} : z \in Z_{1}} = {\overline{α (z)} : z \in Z_{1}}$ .

Facts & Assumptions

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

[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]

The quotient group $G / N$ has the left cosets $g N$ as elements with product $(g N) (h N) : = g h N$ (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

[F3]

For a group homomorphism $f : G \to H$ , $\ker f : = {g \in G : f (g) = e_{H}}$ and $im f : = {f (g) : g \in G}$ (The kernel and image of a group homomorphism).

[L1]

The subgroup $N = {(z, α (z)^{- 1}) : z \in Z_{1}}$ of $G \times H$ is central, hence normal (The identified subgroup used to form a central product is central, hence normal).

[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]

$⟨ S ⟩$ is the smallest subgroup of $G$ containing $S$ , namely $⋂ {K : K \leq G and S \subseteq K}$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L4]

An isomorphism is a bijective group homomorphism (Group isomorphisms, automorphisms and the set $Aut (G)$ ).

Proof

technique · direct

1.1F1F2L1L2

The coordinate maps $g \mapsto (g, e)$ and $h \mapsto (e, h)$ are homomorphisms into $G \times H$ , because the operation there is componentwise, and $π$ is a homomorphism onto the quotient; so both canonical maps are homomorphisms.

2.1F1F3L4step 1.1

The kernel of $g \mapsto \overset{ˉ}{g}$ is ${g \in G : (g, e) \in N}$ ; an equality $(g, e) = (z, α (z)^{- 1})$ gives $g = z$ and $α (z) = e$ , so $z = e$ because $α$ is injective, and the kernel is trivial. Likewise $(e, h) = (z, α (z)^{- 1})$ gives $z = e$ and then $h = α (e)^{- 1} = e$ . Both canonical maps are therefore injective.

2.2L2step 1.1

In $G \times H$ one has $(g, e) (e, h) = (g, h) = (e, h) (g, e)$ , so $\overset{ˉ}{g} \overset{ˉ}{h} = \overset{ˉ}{h} \overset{ˉ}{g}$ for all $g \in G$ and $h \in H$ : the two images commute elementwise.

2.3F2L2L3step 1.1

Every element of $G \circ_{α} H$ is $π (g, h) = π ((g, e) (e, h)) = \overset{ˉ}{g} \overset{ˉ}{h}$ , so the two images together generate $G \circ_{α} H$ .

3.1F1step 2.1step 2.2∎

If $\overset{ˉ}{g} = \overset{ˉ}{h}$ then $(g, e) (e, h)^{- 1} = (g, h^{- 1})$ lies in $N$ , so $g = z \in Z_{1}$ and $h^{- 1} = α (z)^{- 1}$ , that is $h = α (g)$ ; conversely $\overset{ˉ}{z} = \overline{α (z)}$ for every $z \in Z_{1}$ , since $(z, α (z)^{- 1}) \in N$ . Hence the two images meet exactly in ${\overset{ˉ}{z} : z \in Z_{1}}$ .

Remarks

Injectivity is what makes the central product an honest amalgam: each factor embeds, and the only collapsing is the prescribed identification of $Z_{1}$ with $Z_{2}$ . If $α$ were merely a surjective homomorphism, $N$ would meet the first coordinate copy of $G$ in $\ker α \times 1$ and that copy would not embed.

Depends on

Used by

Dependency tree · two levels

18 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

D. A. Craven, The Theory of p-Groups, Proposition 3.5 and Theorem 3.6 (standard reference, not scraped)
M. van Beek, Topics in Finite p-Groups, Definition 2.34 (standard reference, not scraped)