Schoolgirls do not Necessarily Form a Group

In the previous blog, I said that I was not certain that defining an addition on the schoolgirls of the Kirkman Schoolgirl Problem would produce an Abelian group. The Kirkman Schoolgirl problem was to find a 7-day schedule of 3x5 schoolgirl formations among 15 schoolgirls so that each pair of girls marches in the same line once and only once. I then defined an operation + on {0, the schoolgirls} by:

0+0=0
0+a schoolgirl is that same schoolgirl back
Any schoolgirl + herself is 0
If A and B are any two schoolgirls, find the line that contains them (guaranteed by the conditions of the problem). There are three girls in that line, namely A, B, and another schoolgirl C. Define A+B = C.

The operation is defined to be commutative. Also any two schoolgirls added make another one, 0 is the identity, and each girl is her inverse. The only requirement left to make the schoolgirls and 0 into an Abelian group is the associative law:

(A + B) + C = A + (B + C)

I remarked on how hard it would be to prove this on the schoolgirls.

Today I found that the conjecture is false. Given a solution to the schoolgirls problem, if you define + like this, it may not be an Abelian group, since it may not obey the associative law. Take this situation, for example:

	1pm	2pm	3pm	4pm	5pm
Day1	ABC	DEF	GHI	JKL	MNO
Day2	ADG	BEJ	CFM	HKN	ILO
Day3	AEN	BDO	CHL	FIK	GJM
Day4	AIM	BGL	CDK	EHO	FJN
Day5	AHJ	BKM	CEI	DLN	FGO
Day6	AFL	BIN	CJO	DHM	EGK
Day7	AKO	BFH	CGN	DIJ	ELM

This comes from a column on the Mathematical Association of America website. Consider the sums (B + D) + A and B + (D + A). The first one is

(B + D) + A = O + A

This is because BDO form a line in the Kirkman schedule. Further,

O + A = K

Since this OAK tree is in the Kirkman Schedule. So (B + D) + A = K.

The second one runs:

B + (D + A) = B + G

since ADG is a line. Further,

B + G = L,

Since BGL is a line. Therefore,

(B + D) + A != B + (D + A)

So that + is non-associative.

This shows that there is a solution to the Kirkman problem that is not the standard one given by using Klein triples in Z₂⁴. In fact, I hear tell that there are 7 solutions, of which the Z₂⁴ is just one.