# 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

 $f:\bigoplus_{i\in I}G_{i}\to\bigoplus_{i\in I}H_{i}$

is if there exists a family of homomorphisms $\{f_{i}:G_{i}\to H_{i}\}_{i\in I}$ such that

 $f=\bigoplus_{i\in I}f_{i}.$

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

 $f:\bigoplus_{i\in I}G_{i}\to\bigoplus_{i\in I}H_{i}$

is decomposable if and only if for any $j\in I$ and any $g\in\bigoplus_{i\in I}G_{i}$ 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

 $\bigg{(}\bigoplus_{i\in I}f_{i}\bigg{)}^{-1}=\bigoplus_{i\in I}f_{i}^{-1}.$

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

 $\bigg{(}\bigoplus_{i\in I}f_{i}\bigg{)}\circ\bigg{(}\bigoplus_{i\in I}g_{i}% \bigg{)}=\bigoplus_{i\in I}f_{i}\circ g_{i}.$

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

 $f:\bigoplus_{i\in I}G_{i}\to\bigoplus_{i\in I}G_{i}$

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

 $f:\bigoplus_{p\in\mathcal{P}}\mathbb{Z}_{p}\to\bigoplus_{p\in\mathcal{P}}% \mathbb{Z}_{p}$

be a group homomorphism. Then, for any $q\in\mathcal{P}$ and $a\in\bigoplus_{p\in\mathcal{P}}\mathbb{Z}_{p}$ 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 DecomposableHomomorphismsAndFullFamiliesOfGroups 2013-03-22 18:36:03 2013-03-22 18:36:03 joking (16130) joking (16130) 7 joking (16130) Definition msc 20A99