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

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Order, centre and derived subgroup of a central product

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 write $\overset{ˉ}{g}, \overset{ˉ}{h}$ for the canonical images in $P = G \circ_{α} H$ . Then:

if $G$ and $H$ are finite, $∣ P ∣ = \frac{∣ G ∣ ∣ H ∣}{∣ Z_{1} ∣}$ ;
$Z (P) = {\overset{ˉ}{g} \overset{ˉ}{h} : g \in Z (G), h \in Z (H)}$ , the image of $Z (G) \times Z (H)$ ;
$[P, P] = {\overline{c} \overline{d} : c \in [G, G], d \in [H, H]}$ , the image of $[G, G] \times [H, H]$ .

Facts & Assumptions

Given: Groups $G, H$ , central subgroups $Z_{1} \leq Z (G)$ and $Z_{2} \leq Z (H)$ , an isomorphism $α : Z_{1} \to Z_{2}$ , the quotient map $π : G \times H \to P$ , and the canonical images $\overset{ˉ}{g} = π (g, e)$ , $\overset{ˉ}{h} = π (e, h)$ .

[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]

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

[F3]

For $g, h \in G$ the commutator is $[g, h] : = g h g^{- 1} h^{- 1}$ , and $[G, G]$ is the subgroup generated by all commutators (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ ).

[L1]

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 $Z_{1}$ (The two canonical maps into a central product are injective homomorphisms whose images commute, generate it, and meet in the identified centre).

[L2]

If $G$ and $H$ are finite groups, then their external direct product is finite and has order $∣ G \times H ∣ = ∣ G ∣ ∣ H ∣$ (For finite groups $G$ and $H$ , $∣ G \times H ∣ = ∣ G ∣ ∣ H ∣$ ).

[L3]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L4]

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

[L5]

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).

Proof

technique · direct

1.1F1L2L3L5algebra

The map $z \mapsto (z, α (z)^{- 1})$ is a bijection from $Z_{1}$ onto $N$ , so $∣ N ∣ = ∣ Z_{1} ∣$ ; with $∣ G \times H ∣ = ∣ G ∣ ∣ H ∣$ and Lagrange applied to $N \leq G \times H$ , the number of cosets is $∣ G ∣ ∣ H ∣ / ∣ Z_{1} ∣$ , which is the order of $P$ .

1.2L1

The images of $G$ and of $H$ in $P$ commute elementwise, generate $P$ , and each canonical map is injective.

2.1F2F3step 1.2

Let $\overset{ˉ}{g} \overset{ˉ}{h}$ be central in $P$ and let $g^{'} \in G$ . Then $\overset{ˉ}{h}$ commutes with $\overline{g^{'}}$ , so $\overset{ˉ}{g}$ commutes with $\overline{g^{'}}$ , hence $\overline{[g, g^{'}]} = \overline{g} \overline{g^{'}} {\overline{g}}^{- 1} {\overline{g^{'}}}^{- 1}$ is trivial; injectivity of the canonical map gives $[g, g^{'}] = e$ , so $g \in Z (G)$ , and symmetrically $h \in Z (H)$ .

2.2F2step 1.2

Conversely, if $g \in Z (G)$ and $h \in Z (H)$ then $\overset{ˉ}{g} \overset{ˉ}{h}$ commutes with every $\overline{g^{'}}$ and with every $\overline{h^{'}}$ , and those elements generate $P$ , so $\overset{ˉ}{g} \overset{ˉ}{h}$ is central.

2.3F3step 1.2

Because the two images commute elementwise, $[{\overset{ˉ}{g}}_{1} {\overset{ˉ}{h}}_{1}, {\overset{ˉ}{g}}_{2} {\overset{ˉ}{h}}_{2}] = [{\overset{ˉ}{g}}_{1}, {\overset{ˉ}{g}}_{2}] [{\overset{ˉ}{h}}_{1}, {\overset{ˉ}{h}}_{2}] = \overline{[g_{1}, g_{2}]} \overline{[h_{1}, h_{2}]}$ for all $g_{i} \in G$ and $h_{i} \in H$ .

3.1F3L4step 1.1step 2.1step 2.2step 2.3∎

Every element of $P$ is $\overset{ˉ}{g} \overset{ˉ}{h}$ by step 1.2, so steps 2.1 and 2.2 identify $Z (P)$ as the image of $Z (G) \times Z (H)$ ; and by step 2.3 the commutators of $P$ are exactly the products $\overline{[g_{1}, g_{2}]} \overline{[h_{1}, h_{2}]}$ , whose generated subgroup is the image of $[G, G] \times [H, H]$ .

Remarks

Clause 1 needs both factors finite; clauses 2 and 3 do not, since they use only that the two images commute and generate. The order formula divides by $∣ Z_{1} ∣$ and not by $∣ Z_{1} ∣^{2}$ : one copy of the identified subgroup survives inside the product, and it is the image described in the intersection clause of the canonical-maps proposition.

Depends on

Used by

Dependency tree · two levels

31 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 and Proposition 2.36 (standard reference, not scraped)