Memoirs presented to the Cambridge philosophical society on the occasion of the jubilee of Sir George Gabriel Stokes, bart., Hon. LL. D., Hon. SC. D., Lucasian professor.

274 PROF. BURNSIDE, ON A CLASS OF GROUPS OF FINITE ORDER.
These conditions are obviously inconsistent. Hence I2 does exist, and
m + m2=2 {k(2- 1)+ 1l}.
It follows that, ml and m2 being positive numbers of which %n is the greater,
ml > 2k + 1 - k.
On the other hand, since no two operations of order two contained in I, are permutable,
while G contains only 2'2k + 1 subgroups of order 21,
m n 2nk + 1.
Hence there must be an integer 1, less than k, such that
mn1 = 24k + 1 - l,
and m2 = 2'k + 1 + 1- 2k.
Now ml and m2 are relatively prime factors of
(2 + 1)(2n- 1).
Hence (2nk + 1)2 - 2k (2nk + 1) + 2kl - 12 < (2n2k + 1) (2n - 1),
and à fortiori since i is less than k, and 2k + 1 is positive,
2nkc + 1- 2k < 2n- 1,
i.e. k 1.
The group G can therefore only exist if k is unity, and this necessarily involves that
i is zero. Hence
NV= (2n + 1) 2n (2- _ 1), m= 2 + 1, m= 2- -1,
and these are the only values of iV, ry,, and m, consistent with the existence of a
group G having the required property.
Since G is simple, it can be represented as a substitution group of degree 2 + 1.
The subgroup of degree 2n, which leaves one symbol unchanged, has a self-conjugate
Abelian subgroup of order 2n, and 2n conjugate Abelian subgroups of order 2 - 1;
the latter having no common substitutions except identity.
Hence the subgroup of G which leaves one symbol unchanged is doubly-transitive
in the remaining 2" symbols; and therefore G can be represented as a triply-transitive
group of degree 21 + 1.
The Abelian subgroup of order 2 - 1 which transforms a subgroup of degree 2n
is shewn in an appended note to be cyclical. Assuming for the present this result,
the subgroups of G of order 2 (2n-1) are doubly-transitive groups of known type.
Now G contains just n-l1 operations of order two which transform each operation
of a cyclical subgroup of a c l s p o degree 2 - 1 into its inverse. Since each of these leaves
on]y one symbol unchanged, each must interchange the two symbols left unaltered