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

186

SOLVABLE GROUPS

[CH. VIII

operator of order p contained in G and if J is any Sylow subgroup
of order qa, it is evident that s can be transformed into all its
conjugates under G by means of the operators of J. If each
operator of G is represented by a substitution according to which
this operator transforms the conjugates of s, there results a
substitution group S which is simply isomorphic with G, since
G is simple.
To the subgroup J in G there corresponds a transitive substitution group in S, since J transforms s into all its conjugates. As a transitive abelian group is regular, s has exactly
q0 conjugates under G. Hence the identity is the cross-cut
of any two of the qB Sylow subgroups of order pa contained in
G. That is, if G were simple it would contain (pf"-1)q operators whose orders are powers of p, and hence it could contain
only qa operators whose orders are powers of q. As such a
group could contain only one subgroup of order qa, it could not
be simple. This proves that every group of order paq3 is solvable whenever the Sylow subgroups of orders p" and qf are
abelian.
74. Insolvable Groups of Low Composite Orders. From the
theorems which have been established it follows directly that
every group whose order is less than 60 is solvable. That
there is an insolvable group of order 60 and that this is the lowest
order of a simple group of composite order was observed by
E. Galois. We proceed to prove that there is only one insolvable group of this order. If a group of order 60 contains only
one subgroup of order 5, the corresponding quotient group is
of order 12 and hence the group is solvable. Hence an insolvable group of order 60 must involve 6 conjugate subgroups of
order 5 and must transform them according to a transitive substitution group of degree 6 and order 60. As there is only one
such substitution group,* there is only one insolvable group of
order 60.
* The fact that there could not be more than one such substitution group may
be seen as follows: Such a group contains the group of degree 5 and of order 10,
and hence it involves exactly 15 substitutions of order 2. As none of these can
occur in two subgroups of order 4 the group must contain five subgroups of this
order.