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.


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 $\text{Aut}(t(G))$.

  • $\begingroup$ ... and the same argument works for "End". $\endgroup$ – Martin Brandenburg Jul 1 '11 at 1:41
  • $\begingroup$ This seemed clear to me last week, but there is one detail here which still bothers me. Why can you claim that $\alpha(g)_p = \alpha_p(g_p)$? This would certainly work for a direct sum of invariant subgroups, but I don't see why it works for the product. $\endgroup$ – Iian Smythe Jul 8 '11 at 3:42
  • $\begingroup$ @Iian: I have edited the answer to clarify this point. $\endgroup$ – Neil Strickland Jul 8 '11 at 9:57

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