COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.

220 N. S. Mendelsohn
where it is known that the six matrices appearing on the right generate
SL(3, Z).
The following geometrical corollary then follows. The group generated
by all relations of a finite Desarguesian plane which is generated by four
points is a homomorphic image of the collineation group of the free plane
generated by four points.
Example 2. Complete latin squares. A complete latin square of order n
is an array in which every row and every column is a permutation of n
symbols and such that every ordered pair of symbols appears as a consecutive
pair exactly once in the rows and once in the columns. For example
A B C D
C A D B
B D A C
D C B A
is a complete latin square of order 4.
In [2] B. Gordon has shown that complete latin squares exist for every
even order and in [1] E. N. Gilbert has given a number of special constructions
all for even order. It is known that for n = 3, 5, 7, 9 no complete
latin square exists and it had been conjectured that none exists for any odd
order. An exhaustive search by machine is quite impractical for n Z= 11 so
that a specialized search based on an incomplete mathematical theory may
be of use.
Following Gordon we look for a solution in which the square is the
multiplication table for a group. The problem is then reduced to the following.
Let 81, g2, . . ., g,, be the distinct elements of a group. Is it possible to
arrange them so that the elements gl, gF1gZ, g;lgs, . . ., g,=‘,g,, are all
distinct ?
If the group is of odd order and Abelian it is known to be impossible
for such an arrangement to exist. Hence we look to groups which are
non-abelian. The smallest order of a non-abelian group of odd order is 21.
However, a search through the 21! permutations of the elements of the
group is still impractical. The compromise used was to start the arrangement
of the elements of the group by hand until one gets stuck (usually
after 16 to 18 elements). When this is finished the arrangement was put
into the machine with a back-tracking program to try to alter and complete
the arrangement. This proved eminently successful. On p. 221 is an example
of one of the latin squares of order 21.
Example 3. Commutators. There are a number of combinatorial problems
in which it is important to know whether or not an element of a
commutator subgroup is a commutator. More generally the following
question is of interest. Given an element A of the commutator subgroup,
Examples of man-machine interaction
An example of a latin square of order 21 without repeated digraphs
(both rows and columns)
A B C D E F G H I J K L M N O P Q R S T U
K A P I S M T R Q F B E J H C O D N L U G
O M L K B R F D S Q H N E C T J P G U A I
P F S A K H J Q E I N R L O G M C U T B D
E D A J C P S L H B T I G F Q N U K M R O
R C F L G S K T O N J A D I E U B P H Q M
C J E B A N M I L D R H S P U F O T G K Q
F N G C R I D O A L P T H U M S E Q K J B
LQBMOCESNKGDUJ’RTAFHP
M H T P N Q I C K E O U R G J L S D B F A
S l K F P O L E R A U Q T M D H G B J N C
Q G I T D K P J M U L F C A R B N S O E H
T E N H M G R K U P D C B L A Q J F I O S
N P M E T L B U C H F K I Q S G A O R D J
B K O Q L J U N D M A S F R P C I H E G T
G L R N F U H A T C Q O K S B I M J D P E
H O J S U E A G P R M B Q D L T K C N I F
DTQUIBOFJGEMPKh’AHLCSR
JRUOHDQPBsCGNTFELIAMK
I U D G Q A C M F T S J O B H K R E P L N
U S H R J T N B G O I P A E K D F M Q C L
find the minimum value k such that A is expressible as a product of k commutators.
This problem, in general, is known to be unsolvable. However, for free
groups it is solvable but the solution is by no means trivial. The author
has studied the problem of a machine program for making the computations.
To test the possible efficiency of the program the author carried
out hand calculations which would imitate the machine’s behaviour.
Naturally, relations were generated in a random order, but a number of them
proved to be very interesting and one of these led to an interesting theorem.
Here are some of the results turned out at random, all referring to free
groups.
(a) In a free group a word of length six or less which lies in the commutator
subgroup is a commutator, e.g.
a-lb-lc-labc = (ba, bc).
221