Every extraspecial $p$-group is an internal central product of nonabelian subgroups of order $p^3$

If $\mid P\mid =p^3$ the one-member family $P_1=P$ is admissible, and $p^3=p^{1+2\cdot 1}$; moreover any $F=⟨x,y⟩$ with $[x,y]\mathrm{\ne }e$ then has order $p^3$ and equals $P$, so $C_P(F)=Z(P)$ has order $p=\mid P\mid \mathrm{/}{p}^2$ and the splitting clause holds.

Assume all three clauses of the theorem hold for every extraspecial $p$-group of order smaller than $\mid P\mid$.

Suppose $A,B\le P$ satisfy $[A,B]=1$, $P=AB$, $A\cap B=Z(P)$ and $\mid A\mid =p^3$. An element of $Z(B)$ commutes with $B$ and, lying in $B$, with $A$, hence with $AB=P$, so $Z(B)\le Z(P)$; and $Z(P)=A\cap B$ lies in $B$ and is central in $P$, so $Z(B)=Z(P)$. The multiplication map $\mu :A\times B\to P$ is a surjective homomorphism. If $\mu (a,b)=e$, then $a=b^{-1}\in A\cap B$, while every $(z,z^{-1})$ with $z\in A\cap B$ lies in the kernel. Thus $\ker \mu$ is the antidiagonal of $A\cap B$ and has order $p$, so $\mid P\mid =p^3\mid B\mid \mathrm{/}p$ and $\mid B\mid =\mid P\mid \mathrm{/}{p}^2$. If $B$ were abelian it would equal $Z(B)=Z(P)$ and $\mid P\mid$ would be $p^3$; so for $\mid P\mid >p^3$ the group $B$ is nonabelian, and $B\mathrm{/}Z(B)=B\mathrm{/}Z(P)$ is a subgroup of the elementary abelian group $V$, hence elementary abelian, so $B$ is extraspecial.

If $[x,y]\mathrm{\ne }e$ then $F=⟨x,y⟩$ is extraspecial of order $p^3$ with $Z(F)=Z(P)$, and $\beta =b_z(\bar{x},\bar{y})$ is a nonzero element of the field ${\mathbb{F}}_p$, hence invertible.

Let $(P_1,\dots ,P_n)$ be an admissible family and fix an index $i$, writing $C_i=⟨P_j:j\mathrm{\ne }i⟩$. Each generator of $C_i$ commutes with every element of $P_i$, and the centraliser of an element is a subgroup, so $[P_i,C_i]=1$ and $C_i\le C_P(P_i)$; then $P_i{C}_i$ is a subgroup containing every member of the family, so $P=P_i{C}_i$; and $P_i\cap C_i\le P_i\cap C_P(P_i)=Z(P_i)=Z(P)$, while $Z(P)=Z(P_j)\le P_j\le C_i$ gives the reverse inclusion. Taking $C_i$ to be a single $P_j$ shows $P_i\cap P_j=Z(P)$.

For $g\in P$ put $s={\beta }^{-1}{b}_z(\bar{g},\bar{y})$ and $r=-{\beta }^{-1}{b}_z(\bar{g},\bar{x})$ and $u=x^s{y}^r$. Bilinearity and alternation give $b_z(\bar{u},\bar{x})=r{b}_z(\bar{y},\bar{x})=-r\beta =b_z(\bar{g},\bar{x})$ and $b_z(\bar{u},\bar{y})=s{b}_z(\bar{x},\bar{y})=s\beta =b_z(\bar{g},\bar{y})$, so $b_z(\overline{g{u}^{-1}},\bar{x})=0$ and $b_z(\overline{g{u}^{-1}},\bar{y})=0$.

Applying step 1.3 with $A=P_i$ and $B=C_i$ gives $Z(C_i)=Z(P)$ and $\mid C_i\mid =\mid P\mid \mathrm{/}{p}^2$; when $n\ge 2$ the subgroup $C_i$ contains the nonabelian $P_j$, so $\mid P\mid >p^3$ and $C_i$ is extraspecial. The members $P_j$ with $j\mathrm{\ne }i$ generate it, commute pairwise and have centre $Z(C_i)$, so they form an admissible family for it; since $C_i$ has smaller order, the induction hypothesis applied to that $(n-1)$-member family gives $\mid C_i\mid =p^{1+2(n-1)}$.

Hence $g{u}^{-1}$ commutes with $x$ and with $y$; the centraliser of $g{u}^{-1}$ is a subgroup containing both, hence contains $F$, so $g{u}^{-1}\in C_P(F)$ and $g=(g{u}^{-1})u\in C_P(F)F$. Therefore $P=F{C}_P(F)$, and $F\cap C_P(F)=Z(F)=Z(P)$, so $F$ and $C_P(F)$ form an internal central product of $P$; step 1.3 with $A=F$ and $B=C_P(F)$ supplies the order of $C_P(F)$, its centre, and that it is extraspecial when $\mid P\mid >p^3$.

If $\mid P\mid >p^3$, choose $x\mathrm{\notin }Z(P)$, which exists because $P$ is nonabelian, and then $y$ with $b_z(\bar{x},\bar{y})\mathrm{\ne }0$, which exists because the radical is trivial; put $F=⟨x,y⟩$ and $C=C_P(F)$. By step 3.1 the group $C$ is extraspecial of order $\mid P\mid \mathrm{/}{p}^2$, so the induction hypothesis gives an admissible family $P_2,\dots ,P_n$ for $C$ with $\mid C\mid =p^{1+2(n-1)}$. Then $F,P_2,\dots ,P_n$ generate $P$, commute pairwise, are nonabelian of order $p^3$ and have centre $Z(C)=Z(P)$, so they form an admissible family for $P$, and $\mid P\mid =p^2\mid C\mid =p^{1+2n}$; with step 1.1 this completes the induction.