A natural question in group theory is to know whether or not the following implication is true:
Of course, such a cancellation property does not hold in general, for example
whereas and are not isomorphic. However, cancellation property turns out to hold for some classes of groups. In his article On cancellation in groups, Hirshon proves that finite groups are cancellation groups, that is to say, for every groups , if with finite, then . Below, we present an elementary proof of a weaker statement due to Vipul Naik:
Theorem: Let be three finite groups. If then .
Proof. For any finite groups and , let denote the number of homomorphisms from to and denote the number of monomorphisms from to . Notice that
Let be three finite groups such that . Then
for any finite group , since . Using , it is easy to deduce that for any finite group by induction on the cardinality of . Hence
Therefore, there exists a monomorphism from to . From , we deduce that and have the same cardinality, so we conclude that and are in fact isomorphic.
Because a classification of finitely-generated abelian groups or of divisible groups is known, it is easy to verify that cancellation property holds also for such groups (the classification of divisible groups was mentionned in a previous note). However, in his book Infinite abelian groups, Kaplansky mentions that the problem is still open for the class of all abelian groups.
We conclude our note by an example where the cancellation property does not hold even if the groups are all finitely-presented.
Let be the following presentation
Clearly, we have the following decomposition:
Now, we set and . Because commutes with and , we deduce that . Then,
Now, let us notice that
hence . Therefore,
Using , it is not difficult to notice that the relation can be written as . Then, we deduce that
so that the relation can be removed from the presentation. Finally,
Therefore, we find another decomposition of as a direct product:
So . To conclude, it is sufficient to prove that and are not isomorphic. Notice that they are both a semi-direct product .
Let be a morphism. It is not difficult to show that a torsion element of has to be conjugated to , so, without loss of generality, we may suppose that for some . Then, for some . Let be the canonical projection sending to and to . If is onto, then so is : because and , we deduce that . On the other hand,
hence or . Therefore, has to divide and we deduce that . We conclude that cannot be an isomorphism.
Finally, we proved that
where and , but and are not isomorphic.
In particular, we deduce that:
Corollary: is not a cancellation group.