작은숲:공책/군이론/Infinite direct product of integer group is not free

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Statement

A group $P=(\left\{f:\mathbb {N} \to \mathbb {Z} \right\},+)\cong {\mathbb {Z} }^{\omega }$ is not a free abelian group.

Proof

State a lemma related with freeness

Lemma. If F is free and C is a countable subgroup, then the quotient F/C is a direct sum of countable group and a free group.

Proof. Fix a basis B of F, and take B₀ be a subset of B containing the expansion of B₀ contains C.

Then

F/C=\mathbb {Z} (B-B_{0})\oplus \mathbb {Z} (B_{0})/C

: the left part is free and the right part is countable.

Recall Lemma : $\mathbb {Z} (B_{0})$ is countable if $B_{0}$ is countable.
Proof. If $B_{0}$ is countable, then we can find a well-ordering of $B_{0}$ as $B_{0}=\{b_{0},b_{1},\cdot \cdot \cdot \}$ . Then $\mathbb {Z} [B_{0}]$ $=\mathbb {Z} [\left\{b_{0}\right\}]\cup \mathbb {Z} [\left\{b_{0},b_{1}\right\}]\cup \cdot \cdot \cdot \mathbb {Z} [\left\{b_{0},\cdot \cdot \cdot ,b_{n}\right\}$ . Each subset of $\mathbb {Z} [B_{0}]$ is countable, so it is also countable.

So the divisible group F/C should be in the countable summand of above lemma and thus countable.

Now suppose the group P is free. Take $X=\left\{f:\mathbb {N} \to \mathbb {Z} ||\{f(n)!=0\}|<\infty \right\}$ (A subset of P with only finite nonzero function values). Since X is countable, the divisible part of P/X would have countable numbers.

The cardinality of P is equal to $2^{\aleph _{0}}$ , the cardinality of basis of P/X is also $2^{\aleph _{0}}$

However, take a sequence $n\to n!\cdot a_{n}$ for arbitrary sequence $(a_{n})\in \mathbb {Z} ^{\aleph _{0}}$ . Then $n!\cdot a_{n}$ is divisible because for all $N\in \mathbb {N}$ , we can take $(b_{n})$ for $b_{n}={\begin{cases}0&n<N\\n!a_{n}&n\geq N\end{cases}}$ satistying $b_{n}+C=n!\cdot a_{n}+C$ and $(1/N)b_{n}\in P/C$ . Also, $a_{n}\to n!a_{n}$ is a bijective function of P so $\{n!a_{n}\}+C$ has cardinality $2^{\aleph _{0}}$ , which contradicts the lemma.

Therefore, P is not a free abelian group.

Statement

Proof

See also