# characterization of full families of groups

Proposition^{}. Let $\mathcal{G}={\{{G}_{k}\}}_{k\in I}$ be a family of groups. Then $\mathcal{G}$ is full if and only if for any $i,j\in I$ such that $i\ne j$ we have that any homomorphism^{} $f:{G}_{i}\to {G}_{j}$ is trivial.

Proof. ,,$\Rightarrow $” Assume that $f:{G}_{i}\to {G}_{j}$ is a nontrivial group homomorphism. Then define

$$h:\underset{k\in I}{\oplus}{G}_{k}\to \underset{k\in I}{\oplus}{G}_{k}$$ |

as follows: if $t\in I$ is such that $t\ne i$ and $g\in {\oplus}_{k\in I}{G}_{k}$ is such that $g\in {G}_{t}$, then $h(g)=g$. If $g\in {\oplus}_{k\in I}{G}_{k}$ is such that $g\in {G}_{i}$, then $h(g)(j)=f(g(i))$ and $h(g)(k)=0$ for $k\ne j$. This values uniquely define $h$ and one can easily check that $h$ is not decomposable^{}. $\mathrm{\square}$

,,$\Leftarrow $” Assume that for any $i,j\in I$ such that $i\ne j$ we have that any homomorphism $f:{G}_{i}\to {G}_{j}$ is trivial. Let

$$h:\underset{k\in I}{\oplus}{G}_{k}\to \underset{k\in I}{\oplus}{G}_{k}$$ |

be any homomorphism. Moreover, let $i\in I$ and $g\in {\oplus}_{k\in I}{G}_{k}$ be such that $g\in {G}_{i}$. We wish to show that $h(g)\in {G}_{i}$.

So assume that $h(g)\notin {G}_{i}$. Then there exists $j\ne i$ such that $0\ne h(g)(j)\in {G}_{j}$. Let

$$\pi :\underset{k\in I}{\oplus}{G}_{k}\to {G}_{j}$$ |

be the projection and let

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

be the natural inclusion homomorphism. Then $\pi \circ u:{G}_{i}\to {G}_{j}$ is a nontrivial group homomorphism. Contradiction^{}. $\mathrm{\square}$

Corollary. Assume that ${\{{G}_{k}\}}_{k\in I}$ is a family of nontrivial groups such that ${G}_{i}$ is periodic for each $i\in I$. Moreover assume that for any $i,j\in I$ such that $i\ne j$ and any $g\in {G}_{i}$, $h\in {G}_{j}$ orders $|g|$ and $|h|$ are realitvely prime (which implies that $I$ is countable^{}). Then ${\{{G}_{k}\}}_{k\in I}$ is full.

Proof. Assume that $i\ne j$ and $f:{G}_{i}\to {G}_{j}$ is a group homomorphism. Then $|f(g)|$ divides $|g|$ for any $g\in {G}_{i}$. But $f(g)\in {G}_{j}$, so $|g|$ and $|f(g)|$ are relatively prime. Thus $|f(g)|=1$, so $f(g)=0$. Therefore $f$ is trivial, which (due to proposition) completes^{} the proof. $\mathrm{\square}$

Title | characterization of full families of groups |
---|---|

Canonical name | CharacterizationOfFullFamiliesOfGroups |

Date of creation | 2013-03-22 18:36:08 |

Last modified on | 2013-03-22 18:36:08 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 9 |

Author | joking (16130) |

Entry type | Derivation |

Classification | msc 20A99 |