# full families of Hopfian (co-Hopfian) groups

Let $\{G_{i}\}_{i\in I}$ be a full family of groups. Then each $G_{i}$ is Hopfian (co-Hopfian) if and only if $\bigoplus_{i\in I}G_{i}$ is Hopfian (co-Hopfian).

Proof. ,,$\Rightarrow$” Let

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

Of course since $f$ is surjective (injective), then each $f_{i}$ is surjective (injective). Thus each $f_{i}$ is an isomorphism   , because each $G_{i}$ is Hopfian (co-Hopfian). Therefore $f$ is an isomorphism, because

 $f^{-1}=\bigoplus_{i\in I}f_{i}^{-1}.\ \ \square$

,,$\Leftarrow$” Fix $j\in I$ and assume that $f_{j}:G_{j}\to G_{j}$ is a surjective (injective) homomorphism. For $i\in I$ such that $i\neq j$ define $f_{i}:G_{i}\to G_{i}$ to be any automorphism of $G_{i}$. Then

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

is a surjective (injective) group homomorphism. Since $\bigoplus_{i\in I}G_{i}$ is Hopfian (co-Hopfian) then $\bigoplus_{i\in I}f_{i}$ is an isomorphism. Thus each $f_{i}$ is an isomorphism. In particular $f_{j}$ is an isomorphism, which completes    the proof. $\square$

Example. Let $\mathcal{P}=\{p\in\mathbb{N}\ |\ p\mbox{ is prime}\}$ and $\mathcal{P}_{0}$ be any subset of $\mathcal{P}$. Then

 $\bigoplus_{p\in P_{0}}\mathbb{Z}_{p}$

is both Hopfian and co-Hopfian.

Proof. It is easy to see that $\{\mathbb{Z}_{p}\}_{p\in\mathcal{P}}$ is full, so $\{\mathbb{Z}_{p}\}_{p\in\mathcal{P}_{0}}$ is also full. Moreover for any $p\in P_{0}$ the group $\mathbb{Z}_{p}$ is finite, so both Hopfian and co-Hopfian. Therefore (due to proposition)

 $\bigoplus_{p\in P_{0}}\mathbb{Z}_{p}$

is both Hopfian and co-Hopfian. $\square$

Title full families of Hopfian (co-Hopfian) groups FullFamiliesOfHopfiancoHopfianGroups 2013-03-22 18:36:05 2013-03-22 18:36:05 joking (16130) joking (16130) 7 joking (16130) Theorem msc 20A99