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

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FROBENIUS'S THEOREM

79

Among the operators of A there may be several which have
the same constituent P of order pX. All such operators are the
direct product of P and operators whose orders are divisors
of s, and all the operators of A may be divided into distinct
sets such that each set is composed of all the operators of A
which have the same constituent of order p. We proceed to
prove that the number of operators in the combined sets which
involve all the conjugates of P under G is divisible by s, and
hence that the total number of operators in A is divisible by s.
To prove this fact, we consider all the operators of G which
are commutative with P. These form a subgroup H of order
pXr, and the quotient group of H with respect to the cyclic group
generated by P is of order r. The orders of the operators of
this quotient group which divide s must also divide the highest
common factor (t) of s and r. As the order of this quotient
group involves fewer factors than G does we may assume that
the number of its operators whose orders divide t is kt. Hence
A contains exactly kt operators which have the same constituent P.
The combined sets which involve no operator of order pX\
except the conjugates of P under G, must therefore involve
gkt/(pXr) distinct operators, since P has g/(pXr) conjugates
under G. Since g is divisible by s and r, it follows that gt is
divisible by rs, t being the highest common factor of r and s.
Hence s is a divisor of gkt/r. As s and pX are relatively prime,
s must also be a divisor of gkt/(pXr), and the theorem in question has been proved.
While the number of the operators of G whose orders divide
any divisor n of g is always a multiple of n, it does not follow
that groups exist in which the number of these operators is an
arbitrary multiple of n. For instance, if pa is the highest power
of the prime p which divides g, G contains at least one subgroup
of order p", according to Sylow's theorem. If G contains
only one such subgroup this must be invariant and hence G
involves only p' operators whose orders divide pa.
If G contains more than one subgroup of order pa, it must
contain at least p+1 such subgroups, since one such subgroup