Here is a particular Steiner (2, 4, 13) system; other (2, 4, 13) systems might have properties quite different from this one:

	4-set	2-sets
#1	{a, b, c, d}	{a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {c, d}
#2	{a, e, f, g}	{a, e}, {a, f}, {a, g}, {e, f}, {e, g}, {f, g}
#3	{a, h, i, j}	{a, h}, {a, i}, {a, j}, {h, i}, {h, j}, {i, j}
#4	{a, k, l, m}	{a, k}, {a, l}, {a, m}, {k, l}, {k, m}, {l, m}
#5	{b, e, h, k}	{b, e}, {b, h}, {b, k}, {e, h}, {e, k}, {h, k}
#6	{b, f, i, l}	{b, f}, {b, i}, {b, l}, {f, i}, {f, l}, {i, l}
#7	{b, g, j, m}	{b, g}, {b, j}, {b, m}, {g, j}, {g, m}, {j, m}
#8	{c, e, i, m}	{c, e}, {c, i}, {c, m}, {e, i}, {e, m}, {i, m}
#9	{c, f, j, k}	{c, f}, {c, j}, {c, k}, {f, j}, {f, k}, {j, k}
#10	{c, g, h, l}	{c, g}, {c, h}, {c, l}, {g, h}, {g, l}, {h, l}
#11	{d, e, j, l}	{d, e}, {d, j}, {d, l}, {e, j}, {e, l}, {j, l}
#12	{d, f, h, m}	{d, f}, {d, h}, {d, m}, {f, h}, {f, m}, {h, m}
#13	{d, g, i, k}	{d, g}, {d, i}, {d, k}, {g, i}, {g, k}, {i, k}

Here is the rule for establishing a binary quasigroup F (p, q) which can be written more briefly as Fpq or pq, from this Steiner system.

• F (p, p) = p from idempotence.
• If p ≠ q, then find the 4-set of which p and q are both members. F (p, q) will equal some other member of the 4-set. Sometimes an arbitrary choice can be made, but once established it will likely constrain the output of F for other inputs — this is because of cancellativity.

Because the numbers 13 and 4 are relatively prime, there is little prospect of forming a Kirkman-type partition.

What follows is a detailed example of constructing a quasigroup.

The Cayley table for the operation to be defined has 13 rows and 13 columns, but a 4-by-4 sub-table is enough to demonstrate implementation of 4-set #1, namely {a, b, c, d}. Other 4-sets will be handled similarly.

In version 1, pp = p because of the idempotence requirement. Each of the other invocations lists the two possible outputs.

version 1
aa = a	ab = c or d	ac = b or d	ad = b or c	⋯
ba = c or d	bb = b	bc = a or d	bd = a or c	⋯
ca = b or d	cb = a or d	cc = c	cd = a or b	⋯
da = b or c	db = a or c	dc = a or b	dd = d	⋯
⋮	⋮	⋮	⋮	⋱

In version 2, assume without less of generality that ab = c. Then by cancellativity, ad = b and ac = d.

version 2
aa = a	ab = c	ac = d	ad = b	⋯
ba = c or d	bb = b	bc = a or d	bd = a or c	⋯
ca = b or d	cb = a or d	cc = c	cd = a or b	⋯
da = b or c	db = a or c	dc = a or b	dd = d	⋯
⋮	⋮	⋮	⋮	⋱

In version 3, cancellativity is reckoned by columns :

• column 2, db = a and cb = d
• column 3, bc = a and dc = b
• column 4, cd = a and bd = c

version 3
aa = a	ab = c	ac = d	ad = b	⋯
ba = c or d	bb = b	bc = a	bd = c	⋯
ca = b or d	cb = d	cc = c	cd = a	⋯
da = b or c	db = a	dc = b	dd = d	⋯
⋮	⋮	⋮	⋮	⋱

In version 4, cancellativity is reckoned by rows to finish the sub-table:

version 4
aa = a	ab = c	ac = d	ad = b	⋯
ba = d	bb = b	bc = a	bd = c	⋯
ca = b	cb = d	cc = c	cd = a	⋯
da = c	db = a	dc = b	dd = d	⋯
⋮	⋮	⋮	⋮	⋱

Note the ex-commutativity: if p ≠ q, then F (p, q) ≠ F (q, p). The operation, when its definition is complete, will continue to exhibit this characteristic.

(Incidentally, had the original Steiner system been (2, 4, 4), the process would be complete at this point.)

Now for 4-set #2, which is {a, e, f, g}. A 7-by-7 portion of the Cayley table will be used, with irrelevant portions shaded in gray. Three further values are settled by idempotence:

version 5
aa = a	ab = c	ac = d	ad = b	ae = f or g	af = e or g	ag = e or f	⋯
ba = d	bb = b	bc = a	bd = c	be =	bf =	bg =	⋯
ca = b	cb = d	cc = c	cd = a	ce =	cf =	cg =	⋯
da = c	db = a	dc = b	dd = d	de =	df =	dg =	⋯
ea = f or g	eb =	ec =	ed =	ee = e	ef = a or g	eg = a or f	⋯
fa = e or g	fb =	fc =	fd =	fe = a or g	ff = f	fg = a or e	⋯
ga = e or f	gb =	gc =	gd =	ge = a or f	gf = a or e	gg = g	⋯
⋮	⋮	⋮	⋮	⋮	⋮	⋮	⋱

Without loss of generality, choose ae = f. With this Steiner system, a similar arbitrary choice can be made for each of the 13 4-sets, as they all happen to be independent. As a result, 2¹³ = 8192 different (but not especially different) operations can be formed.

With ae = f, the first row can be completed much as in version 2 above.

version 6
aa = a	ab = c	ac = d	ad = b	ae = f	af = g	ag = e	⋯
ba = d	bb = b	bc = a	bd = c	be =	bf =	bg =	⋯
ca = b	cb = d	cc = c	cd = a	ce =	cf =	cg =	⋯
da = c	db = a	dc = b	dd = d	de =	df =	dg =	⋯
ea = f or g	eb =	ec =	ed =	ee = e	ef = a or g	eg = a or f	⋯
fa = e or g	fb =	fc =	fd =	fe = a or g	ff = f	fg = a or e	⋯
ga = e or f	gb =	gc =	gd =	ge = a or f	gf = a or e	gg = g	⋯
⋮	⋮	⋮	⋮	⋮	⋮	⋮	⋱

The rightmost three columns can now be completed as in version 3. Then the bottommost three rows can be completed as in version 4.

version 7
aa = a	ab = c	ac = d	ad = b	ae = f	af = g	ag = e	⋯
ba = d	bb = b	bc = a	bd = c	be =	bf =	bg =	⋯
ca = b	cb = d	cc = c	cd = a	ce =	cf =	cg =	⋯
da = c	db = a	dc = b	dd = d	de =	df =	dg =	⋯
ea = g	eb =	ec =	ed =	ee = e	ef = a	eg = f	⋯
fa = e	fb =	fc =	fd =	fe = g	ff = f	fg = a	⋯
ga = f	gb =	gc =	gd =	ge = a	gf = e	gg = g	⋯
⋮	⋮	⋮	⋮	⋮	⋮	⋮	⋱

The same procedure can be followed for each of the remaining 11 4-sets, and then the operation will be fully defined. The results are below.

complete
aa=a	ab=c	ac=d	ad=b	ae=f	af=g	ag=e	ah=i	ai=j	aj=h	ak=l	al=m	am=k
ba=d	bb=b	bc=a	bd=c	be=h	bf=i	bg=j	bh=k	bi=l	bj=m	bk=e	bl=f	bm=g
ca=b	cb=d	cc=c	cd=a	ce=i	cf=j	cg=h	ch=l	ci=m	cj=k	ck=f	cl=g	cm=e
da=c	db=a	dc=b	dd=d	de=j	df=h	dg=i	dh=m	di=k	dj=l	dk=g	dl=e	dm=f
ea=g	eb=k	ec=m	ed=l	ee=e	ef=a	eg=f	eh=b	ei=c	ej=d	ek=h	el=j	em=i
fa=e	fb=l	fc=k	fd=m	fe=g	ff=f	fg=a	fh=d	fi=b	fj=c	fk=j	fl=i	fm=h
ga=f	gb=m	gc=l	gd=k	ge=a	gf=e	gg=g	gh=c	gi=d	gj=b	gk=i	gl=h	gm=j
ha=j	hb=e	hc=g	hd=f	he=k	hf=m	hg=l	hh=h	hi=a	hj=i	hk=b	hl=c	hm=d
ia=h	ib=f	ic=e	id=g	ie=m	if=l	ig=k	ih=j	ii=i	ij=a	ik=d	il=b	im=c
ja=i	jb=g	jc=f	jd=e	je=l	jf=k	jg=m	jh=a	ji=h	jj=j	jk=c	jl=d	jm=b
ka=m	kb=h	kc=j	kd=i	ke=b	kf=c	kg=d	kh=e	ki=g	kj=f	kk=k	kl=a	km=l
la=k	lb=i	lc=h	ld=j	le=d	lf=b	lg=c	lh=g	li=f	lj=e	lk=m	ll=l	lm=a
ma=l	mb=j	mc=i	md=h	me=c	mf=d	mg=b	mh=f	mi=e	mj=g	mk=a	ml=k	mm=m

Here are the 13 arbitrary choices made for the table above; again, the alternatives would have been equally valid:

	4-set	choice
#1	{a, b, c, d}	ab = c not d
#2	{a, e, f, g}	ae = f not g
#3	{a, h, i, j}	ah = i not j
#4	{a, k, l, m}	ak = l not m
#5	{b, e, h, k}	be = h not k
#6	{b, f, i, l}	bf = i not l
#7	{b, g, j, m}	bg = j not m
#8	{c, e, i, m}	ce = i not m
#9	{c, f, j, k}	cf = j not k
#10	{c, g, h, l}	cg = h not l
#11	{d, e, j, l}	de = j not l
#12	{d, f, h, m}	df = h not m
#13	{d, g, i, k}	dg = i not k

The next matter is to extract a smaller system. From the Steiner (2, 4, 13) system at the top of the page, remove any 4-set and 2-set that does not contain m:

	4-set	2-sets
#4	{a, k, l, m}	{a, m}, {k, m}, {l, m}
#7	{b, g, j, m}	{b, m}, {g, m}, {j, m}
#8	{c, e, i, m}	{c, m}, {e, m}, {i, m}
#12	{d, f, h, m}	{d, m}, {f, m}, {h, m}

Now remove every instance of m, yielding a Steiner (1, 3, 12) system:

	3-set	1-sets
#4	{a, k, l}	{a}, {k}, {l}
#7	{b, g, j}	{b}, {g}, {j}
#8	{c, e, i}	{c}, {e}, {i}
#12	{d, f, h}	{d}, {f}, {h}

A unary quasigroup (in other words, permutation) named P can now be formed. With arbitrary choices resembling those employed above, here is one possible manifestation:

P (a) = k	P (k) = l	P (l) = a
P (b) = g	P (g) = j	P (j) = b
P (c) = e	P (e) = i	P (i) = c
P (d) = f	P (f) = h	P (h) = d