of the Groups ivhose Degree is less than Eight. 225 
and H\ We proceed to prove that H and H' are charac- 
teristic subgroups of G. Suppose that H" were another 
invariant subgroup o£ G which could correspond to H in a 
holomorphism. Hence H" could be constructed by esta- 
blishing some isomorphism between H and W or between 
invariant subgroups. This isomorphism could not be (a, 1), 
since H does not contain any invariant operator besides the 
identity. If it were (a, /3), a, /3>1, the quotient group with 
respect to the subgroup of order a/3 would be composed of 
invariant operators under G. Hence H'' could not be a 
complete group, as its constituents would admit outer iso- 
morphisms under H and H' separately, since these groups do 
not contain any complete group invariantly. From this it 
follows that H must correspond either to itself or to H' in 
every holomorphism of G. As the holomorph of H is the 
square of H, it follows that this holomorph contains all the 
operators of the I of G which transform H into itself. Hence 
this I is the double holomorph * of G. The preceding proof 
clearly also establishes the theorem : The direct product of 
two distinct complete groups neither of which is a direct product 
is a complete group. 
Theorem III. — The group of isomorphisms of the group 
obtained by extending a cyclic group of order 2m, m>2, by 
means of an operator of order 4 which transforms each operator 
of the cyclic group into its inverse is the holomorph of the cyclic 
group. 
The proof of this theorem is similar io that of Theorem I. 
Theorem IV. — If a complete group has only one subgroup 
of index 2, the direct product formed ivith it and the group of 
order 2 is simply isomorphic with its group of isomorphisms. 
Corollary I. The direct product of the symmetric group of 
order n, n zfi 6, and the group of order 2 is simply isomorphic 
with its I. 
Corollary II. The direct product of the metacyclic group of 
order p (p — 1), p being any odd prime, and the group of 
order 2 is simply isomorphic with its I. 
The proof of this theorem follows from the fact that such 
a direct product contains as a characteristic subgroup the 
subgroup of index 2 under the complete group, for if a group 
contains only one subgroup of half its order this subgroup 
is generated by its operators which are squares ; and, vice 
versa, if a subgroup of index 2 is generated by operators 
which are squares, it is the only subgroup of this index. This 
* Transactions of the American Mathematical Society, vol. iv. (1903) 
p. 154. 
