To go further with homology, we will need some additional theory of abelian groups.
We will almost always use additive notation for abelian groups, so the group operation will be denoted by , the identity element by , and the inverse of by . The product of groups and will usually be written rather than .
The basic examples of finitely generated abelian groups are and . Recall that the elements of are the cosets . These can be defined for all , but they repeat with period , so is a complete list of elements.
Let , and be integers such that and is divisible by . Then there is a well-defined homomorphism given by .
We can certainly define a homomorphism by . By assumption we have for some , so , and this is the same as because . This shows that , so we have a well-defined homomorphism as described. ∎
Suppose that and are positive integers that are coprime. Then there is an isomorphism given by
It is clear that the above formula gives a well-defined homomorphism. Next, as and are coprime, we can find integers such that . By the lemma, there are well-defined homomorphisms and given by and . We can combine these to define by . Taking account of the identity , this can be written more explicitly as
From the third and fourth expressions we see that this is equal to mod and equal to mod , so . On the other hand, we have
so is also the identity. ∎
Suppose that , where are distinct primes and . Then there is an isomorphism
given by
This follows from Proposition 12.2 by induction on . ∎
Let be an abelian group, and let be a finite list of elements of . Put
This is easily seen to be a subgroup of . We say that generates if . We say that is finitely generated if there is a finite list that generates it.
The list generates .
For , the standard basis vectors generate .
The list generates .
If generates and generates then the list
generates . Thus, if and are finitely generated, then so is .
We can now see by induction that any group of the form
is finitely generated.
Suppose again that generates . Let be a subgroup of , and let be the usual projection homomorphism, given by . It is then easy to see that the list generates , so is again finitely generated.
If is a finite group then it is certainly finitely generated, because we can just take the full list of elements as generators.
Consider the polynomial ring as an abelian group under addition; we will show that this is not finitely generated. Let be a finite list of elements of , and let be the subgroup of -linear combinations of this list. Let be the maximum of the degrees of all the polynomials . Any element of has the form for some integers , and so has degree at most . It follows that , so is not all of , so does not generate .
Consider the field as an abelian group under addition; we will show that this is not finitely generated. Let be a finite list of elements of , and let be the subgroup of -linear combinations of this list. We can write as for some with . Put , so for all . It follows easily that for all , so , so is not all of , so does not generate .
We now quote two theorems whose proofs can be found in almost any textbook on abstract algebra.
Let be a finitely generated abelian group, and let be a subgroup of . Then is also finitely generated. ∎
Let be a finitely generated abelian group. Then can be expressed (up to isomorphism) as the direct sum of a finite list of summands of the form or (with prime and ). Moreover, the list of summands is unique up to order. ∎
Put and and . We claim that . Indeed, we can define maps
by
It is an exercise to check that these are well-defined and inverse to each other.
Consider an abelian group with . Theorem 12.9 tells us that this can be decomposed as a direct sum of groups of the form , where is prime and . If is a summand, then the order must divide . Thus, the only possible summands are , , , and . The only possibilities for the -power summands are or or . The only possibilities for the -power summands are or . Thus, there are six possibilities for :
One might think that there were additional possibilities like , but the Chinese Remainder Theorem gives , so . Similarly, we have .
The following definition will turn out to be very important.
Consider a pair of homomorphisms of abelian groups. Recall that we have subgroups
We say that the pair is exact if . We also say that the pair is short exact if it is exact, and is injective, and is surjective.
It is standard that is injective iff and is surjective iff . Thus, the sequence is short exact iff we have and and .
For any abelian groups and we have a short exact sequence given by and .
For any we have a short exact sequence given by and .
For any we have a short exact sequence given by and .
If we define by , then the sequence is exact but not short exact.
A sequence is exact iff is injective.
A sequence is exact iff is surjective.
If is exact, then iff is injective.
If is exact, then iff is surjective.
A sequence is exact iff .
Consider a sequence . The image of the first map can only be zero, so the sequence is exact iff , which means that is injective.
Consider a sequence . The kernel of the second map is all of , so the sequence is exact iff , which means that is surjective.
Consider an exact sequence . It is clear that is the zero homomorphism iff is the zero subgroup of , and is injective iff . As and are the same, we see that iff is injective.
Similarly, it is clear that is surjective iff , and iff . As and are the same, we see that is surjective iff .
Consider a sequence . The image of the first map is , and the kernel of the second map is . The sequence is exact iff these two subgroups are the same, iff .
∎
Consider a longer sequence
We say that this is exact at if , so the subsequence
is exact. We say that the whole sequence is exact if it is exact at for all .
Consider a sequence
This is exact at iff is injective, and exact at iff is surjective. Thus, the whole sequence is exact iff is short exact.
Consider the group and the homomorphism given by . We then find that , so the sequence
is exact.
A sequence is exact iff is an isomorphism.
The sequence is exact iff is the image of the map (so ) and is the kernel of the map (which is all of ). The first condition means that is injective, and the second means that is surjective, so the two conditions together mean that is an isomorphism. ∎
Let be a short exact sequence. Then is isomorphic to the subgroup , and the quotient group is isomorphic to . Moreover, the group is finite iff and are both finite, and if so, we have .
As the sequence is assumed to be short exact, we know that is injective and is surjective and .
We can considar as a homomorphism from to . In this context it is surjective (by definition of ) and injective (by assumption) so it is an isomorphism, as required.
Next, as is a surjective homomorphism, the First Isomorphism Theorem tells us that it induces an isomorphism . By the exactness assumption we have , so we have as claimed.
Suppose that and are finite. We can divide into cosets for the subgroup . As is an isomorphism we have , so each coset has size . There is one coset for each element of the group , so the number of cosets is . Thus, the total number of elements is .
Conversely, suppose that is finite. As is isomorphic to a subgroup of , it is also finite. As is isomorphic to a quotient of , it is also finite. We can then go back to the last paragraph to see that again. ∎