Theory and applications of finite groups, by G.A. Miller, H. F. Blichfeldt [and] L. E. Dickson.

116

ABELIAN GROUPS

[CH. IV

product of its Sylow subgroups. If all the subgroups of a Sylow
subgroup of odd order are invariant, this Sylow subgroup must
be abelian. In fact, if the order of such a subgroup Pm is p"r,
then each of its operators of order p must be invariant, since the
group of isomorphisms of the cyclic group of order p is of order
p-1. Suppose that Pm contains a cyclic subgroup Ca of order
pa such that Ca involves non-invariant operators.
If this were possible Pm would contain an operator s which
would transform Ca into itself without being commutative with
every operator of C,. As the operators of Ca would also transform into itself the cyclic group S3 generated by s, it follows
that the commutators, involving elements from Ca and Sp,
would be in both of these cyclic groups. These commutators
would therefore be in the central of the group generated by
Ca and Sp. We may suppose that s was so selected that sp is
commutative with every operator of Ca. As the group generated by Ca and this s would contain non-invariant operators
which would not generate the commutator subgroup of order
p, it results that Pm must be abelian.
We have now proved that every Hamilton group is the direct
product of a Hamilton group of order 2m and some abelian group
of odd order. Hence it remains only to determine the possible
Hamilton groups of order 2m. As an instance of such a group
we may cite the quaternion group. We shall first prove that
such a group H cannot involve any operator whose order exceeds 4.
In fact, if H contained an operator s whose order exceeds 4,
we could find a subgroup in H, by the method used above,
which would contain a non-invariant operator which would
not generate the commutator of this subgroup. Hence we can
assume that the order of every operator of H is either 2 or 4.
Moreover, the operators of order 2 are contained in the central
of H, while each operator of order 4 is transformed either into
itself or into its inverse by every operator of H. Two of
these non-commutative operators of order 4 must have a common square, and hence any two such operators must generate
the quaternion group.