62 A PROBLEM IN PARTITIONS. [722 
The question is important from its connexion with the theory of groups, but it 
seems to be a very difficult one. 
I take the opportunity of mentioning the following theorem : two non-commutative 
symbols a, /3, which are such that /3a — a 2 /3 2 cannot give rise to a group made up of 
symbols of the form a p /3 q . In fact, the assumed relation gives /3a 2 = a 2 /3a 2 /3 2 ; and 
hence, if /3a 2 be of the form in question, = a x [3 y suppose, we have 
a x /3y = a 2 . a x /3 y . /3 2 , = a x+ -(3 y+2 ; 
that is, l=a 2 /3 2 , and thence /3a = 1, that is, /3 = a~ 1 , viz. the symbols are commutative, 
and the only group is that made up of the powers of a.