A Simple Group Having No Multiply Transitive Representation

The totality of 4X4 matrices with coefficients in GF [22], determinant unity, and which permute among themselves the vectors of dimension 4 having 4 co-ordinates nonzero marks of GF [22], or 2 zeros and 2 nonzero marks of the field, forms a simple [l, p. 285, problem 9, part (4)]1 (multiplicative) group, G, of order 26-34-5. It will be demonstrated that G is isomorphic with no multiply transitive permutation group. After submitting the manuscript, the author received a communication, quoted in part, "The referee has observed that if one is willing to use Frame's table (Duke Math. J. vol. 2) for the characters of this group, then the result can be derived in a few lines as follows. The degrees of the irreducible representations are 1, 5, 6, 10, 15, 20, 24, 30,40, 45, 60, 64, 81. If the group had a doubly transitive permutation representation of degree n, then n would divide the order of the group and m —1 would be a degree of an irreducible representation. Moreover, the character of this irreducible representation has to be rational and its values must be ^—1. This eliminates all degrees except n — 1 =15. However, the character in question has the value 3 for an element of order 4 and values —1 for all elements of order 2. This would mean that the element of order 4 leaves 4 letters fixed while its square changes all letters." In the remainder of the paper the counter-example will be established independently of characters: In order that G be representable as at least doubly transitive on n symbols, it is necessary that «! = 26-34-5, and nin—1) be a divisor [l, p. 141, Theorem II] of 26-34-5. Thus n must be 9, 10, 16, or 81. The cases 9 and 10 are easily disposed of, for 10! is divisible by only the fourth power of 3. Hence, if G were represented on 9 or 10 symbols, then each of its Sylow subgroups belonging to 3 must be a Sylow subgroup of the symmetric group; such a subgroup contains a cycle of 3 symbols, which generates a primitive group of degree 3. But a doubly transitive [l, p. 160] group is primitive; and a primitive [3, p. 92, Theorem II] group of degree n containing a primitive subgroup of degree m is at least (« —w + l)-fold transitive. This is the desired contradiction.