Skolem's construction gives a Steiner triple system of order $6m+1$

The even and odd representatives are mapped bijectively by $\pi$ to $A=\{0,\dots ,m-1\}$ and $B=\{m,\dots ,2m-1\}$ respectively. Thus $\pi$ is a permutation. By [F1], for each $x\in Q$ the map $z\mapsto x\circ z$ is a bijection. The operation is commutative, and its diagonal is $a\circ a=a$ and $(m+a)\circ (m+a)=a$ for $0\le a<m$, since the corresponding sums modulo $2m$ are both $2a$. Denote these diagonal values by $d(x)=x\circ x$.

Each listed block has three distinct points. For the third family, $x\mathrm{\ne }y$ makes the two layer-$i$ points distinct, while the last point has layer $i+1\mathrm{\ne }i$. The first and second families plainly have three distinct points. Blocks of the three families cannot coincide: infinity distinguishes the second, and a first-family block uses three layers, while a third-family block uses exactly two. In the third family the layer containing two points uniquely determines $i$ and then $\{x,y\}$, so its indexing produces no duplicate blocks.

A pair containing $\mathrm{\infty }$ occurs only in a second-family block. If its other point is $(m+a,i)$, its unique block is $\{\mathrm{\infty },(m+a,i),(a,i+1)\}$; if its other point is $(a,j)$ with $a\in A$, its unique block is $\{\mathrm{\infty },(m+a,j-1),(a,j)\}$. The two cases partition all finite points and their indices are unique.

A pair of distinct finite points in the same layer, $(x,i),(y,i)$, has $x\mathrm{\ne }y$ and occurs in the unique third-family block indexed by $\{x,y\},i$. Neither of the other families has two finite points in the same layer, and the repeated layer in a third-family block is uniquely determined.

For a finite pair in different layers, there is a unique orientation $((x,i),(y,i+1))$, because the layer set has three elements. A third-family block containing this pair must have repeated layer $i$: a block repeated at $i+1$ uses layers $i+1,i+2$, and one repeated at $i+2$ uses layers $i+2,i$. Such a block therefore has the form $\{(x,i),(z,i),(y,i+1)\}$, where $z\mathrm{\ne }x$ and $x\circ z=y$. Step 1.1 gives a unique solution $z$. It satisfies $z=x$ exactly when $y=d(x)$, so the pair occurs in exactly one third-family block when $y\mathrm{\ne }d(x)$, and in none when $y=d(x)$.

If $y=d(x)$ and $x\in A$, then $y=x$ and the pair lies in the unique first-family block for $x$. If $y=d(x)$ and $x=m+a\in B$, then $y=a$ and the pair lies in the unique second-family block indexed by $a,i$. Conversely, the different-layer finite pairs of every first-family block have the oriented form $(a,i),(a,i+1)$, and the unique finite pair in every second-family block has the oriented form $(m+a,i),(a,i+1)$; these are precisely the two diagonal cases just described. Thus neither family duplicates the third-family pairs from step 2.1, and the diagonal pairs each occur once.

Steps 1.3, 1.4, 2.1 and 3.1 exhaust every pair of distinct points and prove unique block coverage. The point set has $3(2m)+1=6m+1$ points by [F1], and $6m+1\ge 7>3$. Step 1.2 and [F2] now establish the claimed Steiner triple system.