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

148 '

GROUPS DEFINED ABSTRACTLY

[CH. VI

As two operators of order 2 whose product is of order 2 generate the four-group, it results that (si, S2) involves the fourgroup represented by the operators
1, S1SS2 S2S1, S12S2S12.
As this subgroup is invariant under sl as well as under SlS2,
it must be invariant under s2, and therefore also under (si, s2).
Hence the group generated by Si and this subgroup of order
4 is of order 12. Since this group involves Si and Sls2, it must
also involve s2; that is, it must be (si, s2). This proves the
theorem:
If the order of the product of two operators of order 3 is of
order 2 they generate the tetrahedral group.
If we replace Si by ti and sls2 by t2, it results that tl, tl-lt2
are two operators of order 3 whose product is of order 2; hence
(ti, tI-~t2) is the tetrahedral group. Since (ti, tl-lt2)=(ti, 12)
it results also that if the order of the product of two operators
of orders 2 and 3 respectively is 3, then these operators must
generate the tetrahedral group. That is, si, s2 generate the
tetrahedral group if they fulfil either one of the following two
sets of conditions:
l3 =23= (S 2)2 = 1, S1 =S23- (3S2)3 1.
These two sets of equations furnish two very useful definitions
of this important group. The group could also be defined by
the facts that its order is 12 and that it does not involve a subgroup of order 6, as well as by the facts that it is of order 12 and
contains four subgroups of order 3.
The cube is clearly transformed into itself by 24 movements
of rigid space, and the order of each of these movements is
equal to one of the four numbers 1, 2, 3, and 4. It is not difficult to find, among these 24 movements, two of orders 3 and 4
respectively whose product is of order 2. If we represent these
two movements by Si, S2, they must therefore satisfy the equations
S13 S24 = (S1S2)2 =1.