A useful method to prove that two (finite) groups are nonisomorphic is to compare the number of elements of a given order. Of course, the converse is not always true. For example, let us consider Heisenberg group
with a prime number. Noticing that for all and ,
we deduce that is a group of order with elements of order , as . However, and are not isomorphic: one is abelian and not the other.
Nevertheless, it turns out that considering the number of elements of any order is sufficient for finite abelian groups:
Main Theorem: If two finite abelian groups have the same number of elements for any order, then they are isomorphic.
In order to prove our main theorem, we first prove several classical lemmas, usually introduced to show the fundamental theorem of finite abelian groups.
If is a finite abelian group and is a prime number, we define the -torsion of by
It is straightforward to check that is a subgroup of , and our first lemma shows that a finite abelian group is determined by its -torsions.
Lemma 1: Let be a finite abelian group. Then .
Proof. Let and let denote its order. Because are coprime, there exist such that , hence
Let where is a family of different primes. In particular, for some . Therefore, when are large enough,
so either is divisible by or . Thus, .
Finally, we deduce that .
Afterward, we prove Cauchy’s theorem for abelian groups:
Lemma 2: Let be a finite abelian group and be a prime number. If divides , then has an element of order .
Proof. According to lemma 1, it is sufficient to prove that an abelian -group has an element of order . We prove it by induction on . If , then is cyclic and there is nothing to prove. If , either has a nontrivial proper subgroup or is cyclic (and again there is nothing to prove). In the first case, we may apply our induction hypothesis to to deduce that has an element of order .
Lemma 3: Let be a finite abelian -group. If has only one subgroup of order then is cyclic.
Proof. Let us prove lemma 3 by induction on . If then is cyclic and there is nothing to prove. If , let
Because is the set of elements of of order at most , we deduce that . In particular, implies that so that our induction hypothesis may be applied to : there exists an element generating .
Because , we deduce that . But cannot have more than one subgroup of order , hence and finally . Therefore, is cyclic.
Lemma 4: Let be a finite abelian -group and let denote a cyclic subgroup of maximal order. Then contains a subgroup so that .
Proof. Let us prove lemma 4 by induction on . If , and there is nothing to prove. More generally, if is cyclic then and there is nothing to prove, so, if , we may suppose that is not cyclic.
According to lemma 3, has a subgroup of order that is different from the one contained in . In particular, , so is a cyclic subgroup of maximal order in , so that we may apply our induction hypothesis to , and find a subgroup satisfying
Let denote the reciprocal image of in by the quotient map ; in particular, . Because is a subgroup of , . Because in implies in , we deduce that . Therefore, .
Proof of our theorem. Let be two finite abelian groups with the same number of elements of each order. According to lemma 1, we may suppose that they are -groups for some prime . If , clearly .
Now suppose by induction that our theorem holds when one group has an order smaller than or equal to and suppose that . According to lemma 4, there exist , and such that and , where is the highest order of an element of and .
Applying our induction hypothesis, we deduce that hence .
Therefore, our theorem is an easy consequence of lemma 4. It is worth noticing that the fact that any finite abelian group is a direct sum of cyclic groups is also an easy corollary of lemma 4.
We conclude with two corollaries:
Corollary 1: If then .
Corollary 2: If then .
They can be shown by induction on the cardinality and using the identity
where denotes the number of elements of order in some -group.
In fact, the two corollaries above are true for any finite groups.