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

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SYLOW'S THEOREM

29

of G is not divisible by pm+l. Hence there must be at least
one of these double co-sets in which a=3. As pa is the order
of a subgroup of H1, it follows that H1 must contain a Sylow
subgroup. Hence every possible substitution group contains
at least one Sylow subgroup corresponding to every prime number
which divides the order of the group.
If H1 contains more than one subgroup of order pa, let
K1, K2,.. Kx
represent all its subgroups of this order. A substitution s of
K1 which is not also in K2 cannot transform K2 into itself,
otherwise s and K2 would generate a group whose order would
be divisible by pa+1. Hence the substitutions of K1 must
transform K2 into a complete set of conjugates under K1 and
this set contains p"a of these X subgroups. If pal is less than
X-1, this process can be repeated until all of these groups
are exhausted. It is therefore necessary that X-l be divisible
by p. That is, the number of the Sylow subgroups of order pa
contained in any group is always of the form
1+kp.
In other words, this number is iaways -l(mod p).
From this theorem it results directly that every subgroup
of order p7 of a group G is contained in a Sylow subgroup of
G, or is itself a Sylow subgroup of G. In fact, such a subgroup
must transform into itself at least one of the Sylow subgroups
of G. If it were not contained in this Sylow subgroup, G would
involve a subgroup whose order would be a higher power of p
than the order of its Sylow subgroup. It is also clear that
every operator of order pa which is in G and transforms into
itself a Sylow subgroup of order pa must be contained in this
Sylow subgroup.
Another important elementary result in regard to the Sylow
subgroups should be observed here. Suppose that the given
X Sylow subgroups were such that they could not all be transformed into each other by the substitutions of these subgroups.
There would therefore be a set of h which would be transformed
only among themselves by all these substitutions. By trans