8 1. Sylow Theory
(b) Now suppose that n = p is a prime number that divides |G|. Show
that p divides the number of C-orbits of size 1 onO, and deduce
that the number of elements of order p in G is congruent to — 1
Note. In particular, if a prime p divides |G|, then G has at least one element
of order p. This is a theorem of Cauchy, and the proof in this problem is
due to J. H. McKay. Cauchy's theorem can also be derived as a corollary
of Sylow's theorem. Alternatively, a proof of Sylow's theorem different from
Wielandt's can be based on Cauchy's theorem. (See Problem IB.4.)
1A.9. Suppose |G| = pm, where p m and p is prime. Show that G has a
unique subgroup of order p.
1A.10. Let HCG.
(a) Show that
: H\ is equal to the number of right cosets of H
in G that are invariant under right multiplication by H.
(b) Suppose that \H\ is a power of the prime p and that \G : H\ is
divisible by p. Show that
: H\ is divisible by p.
Fix a prime number p. A finite group whose order is a power of p is called a p-
group. It is often convenient, however, to use this nomenclature somewhat
carelessly, and to refer to a group as a "p-group" even if there is no particular
prime p under consideration. For example, in proving some theorem, one
might say: it suffices to check that the result holds for p-groups. What is
meant here, of course, is that it suffices to show that the theorem holds for
all p-groups for all primes p.
We mention that, although in this book a p-group is required to be finite,
it is also possible to define infinite p-groups. The more general definition is
that a (not necessarily finite) group G is a p-group if every element of G has
finite p-power order. Of course, if G is finite, then by Lagrange's theorem,
every element of G has order dividing |G|, and so if |G| is a power of p, it
follows that the order of every element is a power of p, and hence G is a
p-group according to the more general definition. Conversely, if G is finite
and has the property that the order of every element is a power of p, then
clearly, G can have no element of order q for any prime q different from p.
It follows by Cauchy's theorem (Problem 1A.8) that no prime q ^ p can
divide |G|, and thus |G| must be a power of p, and this shows that the two
are equivalent for finite groups.
Again, fix a prime p. A subgroup S of a finite group G is said to be
a Sylow p-subgroup of G if |5| is a power of p and the index \G : S\ is