fundamental theorem of finitely generated abelian groups in nLab

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Group Theory

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• Examples
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Theorem

(fundamental theorem of finitely generated abelian groups)

Every finitely generated abelian group AA is isomorphic to a direct sum of p-primary cyclic groups ℤ/p kℤ\mathbb{Z}/p^k \mathbb{Z} (for pp a prime number and kk a natural number ) and copies of the infinite cyclic group ℤ\mathbb{Z}:

A≃ℤ n⊕⨁iℤ/p i k iℤ. A \;\simeq\; \mathbb{Z}^n \oplus \underset{i}{\bigoplus} \mathbb{Z}/p_i^{k_i} \mathbb{Z} \,.

The summands of the form ℤ/p kℤ\mathbb{Z}/p^k \mathbb{Z} are also called the p-primary components of AA. Notice that the p ip_i need not all be distinct.

fundamental theorem of finite abelian groups:

In particular every finite abelian group is of this form for n=0n = 0, hence is a direct sum of cyclic groups.

fundamental theorem of cyclic groups:

In particular every cyclic group ℤ/nℤ\mathbb{Z}/n\mathbb{Z} is a direct sum of cyclic groups of the form

ℤ/nℤ≃⨁iℤ/p i k iℤ \mathbb{Z}/n\mathbb{Z} \simeq \underset{i}{\bigoplus} \mathbb{Z}/ p_i^{k_i} \mathbb{Z}

where all the p ip_i are distinct and k ik_i is the maximal power of the prime factor p ip_i in the prime decomposition of nn.

Specifically, for each natural number dd dividing nn it contains ℤ/dℤ\mathbb{Z}/d\mathbb{Z} as the subgroup generated by n/d∈ℤ→ℤ/nℤn/d \in \mathbb{Z}\to \mathbb{Z}/n\mathbb{Z}. In fact the lattice of subgroups of ℤ/nℤ\mathbb{Z}/n\mathbb{Z} is the formal dual of the lattice of natural numbers ≤n\leq n ordered by inclusion.

Example

For p 1p_1 and p 2p_2 two distinct prime numbers, p 1≠p 2p_1 \neq p_2, then there is, up to isomorphism, precisely one abelian group of order p 1p 2p_1 p_2, namely

ℤ/p 1ℤ⊕ℤ/p 2ℤ. \mathbb{Z}/p_1 \mathbb{Z} \oplus \mathbb{Z}/p_2 \mathbb{Z} \,.

This is equivalently the cyclic group

ℤ/p 1p 2ℤ≃ℤ/p 1ℤ⊕ℤ/p 2ℤ. \mathbb{Z}/p_1 p_2 \mathbb{Z} \simeq \mathbb{Z}/p_1 \mathbb{Z} \oplus \mathbb{Z}/p_2 \mathbb{Z} \,.

The isomorphism is given by sending 11 to (p 2,p 1)(p_2,p_1).

Example

Moving up, for two distinct prime numbers p 1p_1 and p 2p_2, there are exactly two abelian groups of order p 1 2p 2p_1^2 p_2, namely (ℤ/p 1ℤ)⊕(ℤ/p 1ℤ)⊕(ℤ/p 2ℤ)(\mathbb{Z}/p_1 \mathbb{Z})\oplus (\mathbb{Z}/p_1 \mathbb{Z}) \oplus (\mathbb{Z}/p_2 \mathbb{Z}) and (ℤ/p 1 2ℤ)⊕(ℤ/p 2ℤ)(\mathbb{Z}/p_1^2 \mathbb{Z})\oplus (\mathbb{Z}/p_2 \mathbb{Z}). The latter is the cyclic group of order p 1 2p 2p_1^2 p_2. For instance, ℤ/12ℤ≅(ℤ/4ℤ)⊕(ℤ/3ℤ)\mathbb{Z}/12\mathbb{Z} \cong (\mathbb{Z}/4 \mathbb{Z})\oplus (\mathbb{Z}/3 \mathbb{Z}).

Example

Similarly, there are four abelian groups of order p 1 2p 2 2p_1^2 p_2^2, three abelian groups of order p 1 3p 2p_1^3 p_2, and so on.

More generally, theorem may be used to compute exactly how many abelian groups there are of any finite order nn (up to isomorphism): write down its prime factorization, and then for each prime power p kp^k appearing therein, consider how many ways it can be written as a product of positive powers of pp. That is, each partition of kk yields an abelian group of order p kp^k. Since the choices can be made independently for each pp, the numbers of such partitions for each pp are then multiplied.

Of all these abelian groups of order nn, of course, one of them is the cyclic group of order nn. The fundamental theorem of cyclic groups says it is the one that involves the one-element partitions k=[k]k= [k], i.e. the cyclic groups of order p kp^k for each pp.