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# Baer-Specker group

Let $A$ be a non-empty set, and $G$ an abelian group. The set $K$ of all functions from $A$ to $G$ is an abelian group, with addition defined elementwise by $(f+g)(x)=f(x)+g(x)$. The zero element is the function that sends all elements of $A$ into $0$ of $G$, and the negative of an element $f$ is a function defined by $(-f)(x)=-(f(x))$.

When $A=\mathbb{N}$, the set of natural numbers, and $G=\mathbb{Z}$,
$K$ as defined above is called the *Baer-Specker group*. Any
element of $K$, being a function from $\mathbb{N}$ to $\mathbb{Z}$,
can be expressed as an infinite sequence $(x_{1},x_{2},\ldots,x_{n},\ldots)$, and the elementwise addition on $K$ can
be realized as componentwise addition on the sequences:

$(x_{1},x_{2},\ldots,x_{n},\ldots)+(y_{1},y_{2},\ldots,y_{n},\ldots)=(x_{1}+y_{% 1},x_{2}+y_{2},\ldots,x_{n}+y_{n},\ldots).$ |

An alternative characterization of the Baer-Specker group $K$ is that it can be viewed as the countably infinite direct product of copies of $\mathbb{Z}$:

$K=\mathbb{Z}^{{\mathbb{N}}}\cong\mathbb{Z}^{{\aleph_{0}}}=\prod_{{\aleph_{0}}}% \mathbb{Z}.$ |

The Baer-Specker group is an important example of a torsion-free abelian group whose rank is infinite. It is not a free abelian group, but any of its countable subgroup is free (abelian).

# References

- 1 P. A. Griffith, Infinite Abelian Group Theory, The University of Chicago Press (1970)

## Mathematics Subject Classification

20K20*no label found*

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## Corrections

Finitely many nonzero elements? by lars_h ✓

Proper subgroups by lars_h ✓

continuum hypothesis by yark ✓