Cancellation in finite groups
by David Radcliffe
In this note, I will state a proof of the following theorem: If are finite groups, and if is isomorphic to , then is isomorphic to .
Let be the number of homomorphisms from to , and let be the number of monomorphisms from to .
If is a finite group then
But since and are isomorphic. Therefore . Dividing both sides by yields .
I claim that for every finite group . The proof is by induction on the order of . If has order 1 then and , so the equality holds.
Suppose that for every group of order less than , and let be a group of order .
where ranges over all nontrivial normal subgroups of . But by the preceding argument, and by the induction hypothesis. Therefore .
Setting and in the above yields and . Thus there exist monomorphisms and . Since and are finite, this implies that and are isomorphic. QED.
I am very fond of this theorem because it was a key lemma in my doctoral dissertation. I learned of this argument from Fred Galvin, but I do not know if it was original to him.
The theorem would be false if the word ‘finite’ were omitted. For example, if A is a free abelian group of infinite rank, B is an infinite cyclic group, and C is the trivial group, then and are both isomorphic to A, but B and C are not isomorphic to each other.