# decomposable homomorphisms and full families of groups

Let ${\{{G}_{i}\}}_{i\in I},{\{{H}_{i}\}}_{i\in I}$ be two families of groups (indexed with the same set $I$).

Definition. We will say that a homomorphism^{}

$$f:\underset{i\in I}{\oplus}{G}_{i}\to \underset{i\in I}{\oplus}{H}_{i}$$ |

is decomposable^{} if there exists a family of homomorphisms ${\{{f}_{i}:{G}_{i}\to {H}_{i}\}}_{i\in I}$ such that

$$f=\underset{i\in I}{\oplus}{f}_{i}.$$ |

Remarks. For each $j\in I$ and $g\in {\oplus}_{i\in I}{G}_{i}$ we will say that $g\in {G}_{j}$ if $g(i)=0$ for any $i\ne j$. One can easily show that any homomorphism

$$f:\underset{i\in I}{\oplus}{G}_{i}\to \underset{i\in I}{\oplus}{H}_{i}$$ |

is decomposable if and only if for any $j\in I$ and any $g\in {\oplus}_{i\in I}{G}_{i}$ such that $g\in {G}_{j}$ we have $f(g)\in {H}_{j}$. This implies that if $f$ is an isomorphism^{} and $f$ is decomposable, then each homomorphism in decomposition is an isomorphism and

$${\left(\underset{i\in I}{\oplus}{f}_{i}\right)}^{-1}=\underset{i\in I}{\oplus}{f}_{i}^{-1}.$$ |

Also it is worthy to note that composition^{} of two decomposable homomorphisms is also decomposable and

$$\left(\underset{i\in I}{\oplus}{f}_{i}\right)\circ \left(\underset{i\in I}{\oplus}{g}_{i}\right)=\underset{i\in I}{\oplus}{f}_{i}\circ {g}_{i}.$$ |

Definition. We will say that family of groups ${\{{G}_{i}\}}_{i\in I}$ is full if each homomorphism

$$f:\underset{i\in I}{\oplus}{G}_{i}\to \underset{i\in I}{\oplus}{G}_{i}$$ |

is decomposable.

Remark. It is easy to see that if ${\{{G}_{i}\}}_{i\in I}$ is a full family of groups and ${I}_{0}\subseteq I$, then ${\{{G}_{i}\}}_{i\in {I}_{0}}$ is also a full family of groups.

Example. Let $\mathcal{P}=\{p\in \mathbb{N}|p\text{is prime}\}$. Then ${\{{\mathbb{Z}}_{p}\}}_{p\in \mathcal{P}}$ is full. Indeed, let

$$f:\underset{p\in \mathcal{P}}{\oplus}{\mathbb{Z}}_{p}\to \underset{p\in \mathcal{P}}{\oplus}{\mathbb{Z}}_{p}$$ |

be a group homomorphism. Then, for any $q\in \mathcal{P}$ and $a\in {\oplus}_{p\in \mathcal{P}}{\mathbb{Z}}_{p}$ such that $a\in {\mathbb{Z}}_{q}$ we have that $|a|$ divides $q$ and thus $|f(a)|$ divides $q$, so it is easy to see that $f(a)\in {\mathbb{Z}}_{q}$. Therefore (due to first remark) $f$ is decomposable.

Counterexample. Let ${G}_{1},{G}_{2}$ be two copies of $\mathbb{Z}$. Then $\{{G}_{1},{G}_{2}\}$ is not full. Indeed, let

$$f:\mathbb{Z}\oplus \mathbb{Z}\to \mathbb{Z}\oplus \mathbb{Z}$$ |

be a group homomorphism defined by

$$f(x,y)=(0,x+y).$$ |

Now assume that $f={f}_{1}\oplus {f}_{2}$. Then we have:

$$(0,1)=f(1,0)=({f}_{1}(1),{f}_{2}(0))$$ |

and so ${f}_{2}(0)=1$. Contradiction^{}, since group homomorphisms preserve neutral elements.

Title | decomposable homomorphisms and full families of groups |
---|---|

Canonical name | DecomposableHomomorphismsAndFullFamiliesOfGroups |

Date of creation | 2013-03-22 18:36:03 |

Last modified on | 2013-03-22 18:36:03 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 7 |

Author | joking (16130) |

Entry type | Definition |

Classification | msc 20A99 |