Let $G = \prod\_p \mathbb{Z}/p\mathbb{Z}$, where $p$ ranges over all primes, considered as an abelian group. What does $\text{Aut}(G)$ (or even $\text{End}(G)$) look like?


I know that that if we take $t(G) = \oplus\_p \mathbb{Z}/p\mathbb{Z}$, then $\text{Aut}(t(G)) = \prod\_p (\mathbb{Z}/p\mathbb{Z})^\times$ (can be thought of as infinite diagonal matrices), but I am more interested in the product.


More generally, if $G = \prod\_{i \in I} G\_i$, where each $G\_i$ is fully invariant (embedded in the obvious way), what can we say about $\text{Aut}(G)$?


EDIT: I do apologize for the imprecision of the question, but anything would be helpful, and an exact description, as in the case for the direct sum, would be fantastic.



Automorphisms of an infinite direct product of abelian groups
Put $A\_p=\mathbb{Z}/p$ and $B\_p=\prod\_{q\neq p}A\_q$ so $G=A\_p\times B\_p$. Multiplication by $p$ acts as zero on $A\_p$ and as an automorphism on $B\_p$ so $pG=0\times B\_p$ and $G/pG\simeq A\_p$. Put $U\_p=\text{Aut}(A\_p)=(\mathbb{Z}/p)^\times$. There is an evident homomorphism $\phi:\prod\_pU\_p\to\text{Aut}(G)$. Any automorphism $\alpha\in\text{Aut}(G)$ induces an automorphism $\alpha\_p\in\text{Aut}(G/pG)=\text{Aut}(A\_p)=U\_p$. In more detail, the projection $\pi\_p:G\to A\_p$ is surjective, and $\ker(\pi\_p)=0\times B\_p=pG$, which implies that $\alpha(\ker(\pi\_p))\leq\ker(\pi\_p)$. From this it follows that there is a unique homomorphism $\alpha\_p:A\_p\to A\_p$ with $\alpha\_p\pi\_p=\pi\_p\alpha$, or in other words $\alpha(g)\_p=\alpha\_p(g\_p)$ for all $g\in G$.


We thus have a map $\psi:\text{Aut}(G)\to\prod\_pU\_p$ sending $\alpha$ to $(\alpha\_p)\_{p\in\text{Primes}}$. If $\psi(\alpha)=1$ then for all $g\in G$ we have $\alpha(g)\_p=\alpha\_p(g\_p)=g\_p$ for all $p$, which means that $\alpha(g)=g$; this means that $\psi$ is injective. Given this, it is not hard to check that $\phi$ and $\psi$ are inverse to each other. Thus, $\text{Aut}(G)$ is the same as $\text{Aut}(t(G))$. 


