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

144 GROUPS DEFINED ABSTRACTLY [CH. VI
of the operators si and s2. From the fact that S12=s22, it
results that
S1S2-1 = S1-l2S2 -I =S1-1S2, S2S1 -1 = S2I-lS = (SS2-1)-1.
Each of the operators sls2-l, s2s1-l is therefore transformed
into the other by each of the two operators sl, s2. That is,
the cyclic group generated by either of these operators is
invariant under G, and each of its operators is transformed into
its inverse by sl as well as by s2. The abelian group generated
by the two operators Si2, slS2 - must therefore be invariant
under G, and it involves either all, or just half of the operators
of G.
When each of the two operators sl, s2 is of odd order, G is
cyclic, since each of these operators is equal to the other, and
G is generated by si2 in this case. When one of these operators
is of odd order while the other is of even order, G is generated
by the operator of even order and sls2-l is of order 2. The
only case which requires further consideration is therefore
the one in which the common order of sl, s2 is an even number
2n. A necessary and sufficient condition that G be abelian
is that the order of SiS2-1 divide 2. If sls2-l=l, Si=s2 and
G is the cyclic group generated by sl. If the order of sls2-l
is 2, G is either the cyclic group of order 2n or the abelian group
whose invariants are 2 and 2n. It remains only to consider
the cases when G is non-abelian; that is, when the order of
sis2-~ exceeds 2, and hence sl, s2 have the same even order 2n.
It is evident that the common order of sl, s2 is not limited
by the relation S12=S22. That is, for an arbitrary value of n
we can find two operators sl, s2 of order 2n such that they satisfy
the equation S12 =S22. In fact, the value of n is not limited if
we impose the additional condition that the order of sls2-l
shall be an arbitrary number m, since two generators of order
2 of the dihedral group of order 2m may be multiplied by an
operator of order 2n which is commutative with both of these
generators such that the products satisfy both of these conditions. In other words, for any arbitrary number pair m,
n, we can find two operators si, S2 of order 2n such that they