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

18

SUBSTITUTION GROUPS

[CH. II

words, either all the substitutions of a substitution group are positive or exactly half of them are positive.
Whenever a group contains negative substitutions it contains a subgroup of half its own order, composed of its positive
substitutions. In particular, the symmetric group of degree
n contains a subgroup of order n!/2 which is composed of its
positive substitutions. This subgroup is called the alternating
group of degree n. Hence there are at least three distinct
groups of degree n whenever n>3, viz., the group generated
by a cyclic substitution on n letters, the alternating group, and
the symmetric group. It will be proved that other groups of
degree n exist for every value of n> 3. The cyclic substitution of degree n is positive or negative according as n is odd
or even.
The product of two transpositions which have a common
letter is always of the form abc. A positive cyclic substitution is always the product of substitutions of the form abc,
since it is the product of an even number of transpositions
having a common letter. A substitution composed of two
negative cyclic substitutions is also the product of substitutions of the form abc. In fact, such a substitution may be
regarded as the product of two distinct sets of transpositions
such that all the transpositions of each set have a common
letter and such that each set involves an odd number of transpositions. Hence it remains only to observe that a substitution composed of two transpositions having no letter in common
is the product of substitutions of the form abc. This fact
results directly from the product
ab. cd = acb bdc.
That is, every possible positive substitution is the product of substitutions of the form abc.
8. Commutative Substitutions. Let s and t represent two
substitutions. If st=ts, these two substitutions are said to be
commutative. For instance, if s = ab cd and t = ac bd, it is easy
to verify that st=ts=ad.bc. On the other hand, if s=ab.cd
and tl=bc, it results that st =acdb while t1s=abdc. Hence