Every group besides the abelian group of order 2a and of type (1, 1, 1, ***) contains at least two operators which are squares of other operators in the group. If there are only two such operators they constitute an invariant subgroup and the corresponding quotient group contains only one operator which is a square. All the groups which satisfy this condition have recently been determined.t The present paper is devoted to a complete determination of the groups involving just three operators which are squares. As the idenitity is its own square such a group ( G) contains only two operators besides the identity which have the property in question. If the order (g) of G is divisible by an odd number, this number is 3 and G includes only one subgroup of this order, since every operator of odd order is the square of some power of itself. As suich a group cannot contain an operator of order 4 and its order is 3.2a, it niust be the direct product of the abelian grouip of order 2a and of type (1 , 1, 1, * * *), and the cycle group of order 3; or one of its subgroups of order 3.2a-1 must be of this type. In the latter case, G is evidently the direct product of the symmetric group of order 6 and the abelian group of order 2a-1 and of type ( 1, 1, 1, *) . In what follows, these trivial cases will not be considered. Hence g will always be of the form 2a and G contains only operators of orders two and four in addition to the identity. Moreover, each of the two operators (t,, t2) which are squares but are not the identity is of order two and hence it is the square of some operators of order four. We shall first prove that t1, t2 are invariant under G.