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

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Internal central products are the images of external ones

Statement

Let $G$ be a group and $G_{1}, \dots, G_{r} \leq G$ . Subgroups $G_{1}, \dots, G_{r}$ form an internal central product of $G$ if and only if the multiplication map $G_{1} \times \dots \times G_{r} \to G$ , $(g_{1}, \dots, g_{r}) \mapsto g_{1} \dots g_{r}$ , is a surjective homomorphism; each factor then meets its kernel trivially.

For two factors this identifies the internal notion with the external one: if $G_{1}, G_{2}$ form an internal central product of $G$ and $D = G_{1} \cap G_{2}$ , then $D \leq Z (G_{1})$ and $D \leq Z (G_{2})$ , and

$G ≅ G_{1} \circ_{{id}_{D}} G_{2},$

the external central product of The central product $G \circ_{α} H$ of two groups along an isomorphism of central subgroups taken along the identity isomorphism of $D$ .

Facts & Assumptions

Given: A group $G$ and subgroups $G_{1}, \dots, G_{r} \leq G$ ; in the second half, $r = 2$ and $D = G_{1} \cap G_{2}$ .

[F1]

Subgroups $G_{1}, \dots, G_{r}$ of $G$ form an internal central product when they generate $G$ and $[G_{i}, G_{j}] = 1$ for $i \neq j$ (Internal central products of a finite family of subgroups).

[F2]

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

[F3]

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).

[F4]

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]

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

[L2]

$⟨ S ⟩$ is the smallest subgroup of $G$ containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L3]

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

Proof

technique · direct

1.1

Suppose the subgroups generate $G$ and commute pairwise. Writing $μ (g_{1}, \dots, g_{r}) = g_{1} \dots g_{r}$ , the commuting hypothesis lets the factors of $μ (x) μ (y)$ be sorted by index, so $μ (x y) = μ (x) μ (y)$ and $μ$ is a homomorphism; its image is a subgroup containing every $G_{i}$ , hence equals $⟨ G_{1}, \dots, G_{r} ⟩ = G$ , so $μ$ is surjective.

F1F2F4L2algebra

1.2

For the converse, suppose $μ$ is a surjective homomorphism. Surjectivity gives $G = G_{1} \dots G_{r} \subseteq ⟨ G_{1}, \dots, G_{r} ⟩$ , so the subgroups generate. For $i \neq j$ , $a \in G_{i}$ and $b \in G_{j}$ , the tuples $x$ with $a$ in place $i$ and $y$ with $b$ in place $j$ commute in the direct product, so $a b = μ (x) μ (y) = μ (x y) = μ (y x) = μ (y) μ (x) = b a$ ; hence $[G_{i}, G_{j}] = 1$ .

F1F2L2algebra

1.3

In either case a tuple with a single nonidentity entry $g_{i}$ has $μ$ -value $g_{i}$ , so it lies in $\ker μ$ only if $g_{i} = e$ : each factor meets the kernel trivially.

2.1

Now let $r = 2$ and let $G_{1}, G_{2}$ form an internal central product with $D = G_{1} \cap G_{2}$ . An element $d \in D$ lies in $G_{2}$ , so it commutes with every element of $G_{1}$ , giving $D \leq Z (G_{1})$ ; symmetrically $D \leq Z (G_{2})$ .

F1L3step 1.1

2.2

The kernel of $μ : G_{1} \times G_{2} \to G$ is ${(g_{1}, g_{2}) : g_{1} g_{2} = e} = {(d, d^{- 1}) : d \in D}$ , since $g_{1} = g_{2}^{- 1}$ lies in both subgroups.

F4step 1.1

3.1

That kernel is exactly the subgroup $N$ used to build $G_{1} \circ_{{id}_{D}} G_{2}$ , so the first isomorphism theorem gives $G ≅ (G_{1} \times G_{2}) / N = G_{1} \circ_{{id}_{D}} G_{2}$ .

F3L1step 1.1step 2.1step 2.2∎

Remarks

The commuting condition is imposed only for $i \neq j$ . A single factor is not required to be abelian, which is what allows a nonabelian group to be an internal central product of one factor, namely itself.

The identity isomorphism of $D$ is forced here rather than chosen: the kernel of the multiplication map is ${(d, d^{- 1})}$ , and that is the identified subgroup of the external product along ${id}_{D}$ and along no other map.

Depends on

Used by

Dependency tree · two levels

23 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 (standard reference, not scraped)
M. van Beek, Topics in Finite p-Groups, Definition 2.34 (standard reference, not scraped)