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

114

ABELIAN GROUPS

[CH. IV

tively. We proceed to prove that Pi is cyclic and that P2 is of
type (1, 1, 1,...). That P1 is cyclic follows directly from
the fact that s transforms P2 into itself and that G is generated
by s and P2. In fact, it is evident that ca =3. If P2 were
not of type (1, 1,... ), it would contain characteristic
subgroups generated by its operators whose orders are divisors of p2, p22,.., P2r-1, where p2r is the order of its operators of highest order. All of the operators of these characteristic subgroups would be composed of operators which would
be commutative with s, since G cannot contain a non-abelian
subgroup. Hence s would have to transform among themselves all the operators of order p2r in P2 which have the same
p2th power. As the number of these operators is a power
of p2, this is impossible. That is, we have arrived at an absurdity by assuming that r> 1, and hence we have established the
theorem: If a non-abelian group which contains only abelian
subgroups has more than one Sylow subgroup, one of these subgroups is of the type (1, 1,,... ) and the others are cyclic.
EXERCISES
1. If all the subgroups of a non-abelian group of order pm, p being a
prime number, are abelian, its commutator subgroup is of order p and
the pth power of each of its operators is invariant.*
2. Every subgroup of the dicyclic group of order 4p is abelian.
3. If every subgroup of order pm-l in a non-abelian group of order
pm is abelian, there must be exactly p +1 such subgroups.
4. The group of order 56 which contains 8 subgroups of order 7 does
not involve any non-abelian subgroup.
46. Roots of the Operators of an Abelian Group. If si, s2,., s are the operators of an abelian group G, and if G contains two operators s,, so which are such that s a=s5, then s,
is said to be an nth root of so. In particular, every operator
of G is a gth root of the identity, so that the identity has g gth
roots under G. If n is prime to g every operator of G has one
and only one nth root. On the other hand, if n is a divisor
of g, the total number of the operators of G whose orders divide
* Cf. G. A. Miller and H. C. Moreno, Transactions of the American Mathematical Society, vol. 4 (1903), p. 403.