# When is $G\cong\operatorname{End}(G)$?

$\newcommand\End{\operatorname{End}}$Let $G$ be an Abelian group.

Are there sufficient conditions for the existence of an isomorphism $G\cong\End(G)$, where $\End(G)$ is considered a group under addition? This is true for $\mathbb Z$ and $\mathbb Q$, but not (as far as I know) for $\mathbb R$, where you can construct monstrous additive functions from the Axiom of Choice.

Side questions for people to consider: If $R$ is a ring, when will it be isomorphic (as a ring) to the ring $\End_{\text{group}}(R)$ of group endomorphisms? When will it be isomorphic to the ring $\End_{\text{ring}}(R)$ of ring endomorphisms?

• Regarding the second part of the side question: the ring endomorphisms of a ring don't form a ring. – Jeremy Rickard Apr 11 '15 at 10:09
• With composition as multiplication, I'm quite sure they do. – Gaussler Apr 11 '15 at 10:26
• @Gaussier For $R = ℤ$, $f = \mathrm{id}_ℤ + \mathrm{id}_ℤ$ is not multiplicative (nor does it preserve $1$), because $f(1·1) = 2 ≠ 4 = f(1)·f(1)$ (that’s right – $\mathrm{Ring}$ is not an additive category). – k.stm Apr 11 '15 at 11:11
• Ah, you're right. Damn rings with one! – Gaussler Apr 11 '15 at 11:13
• @Gaussler Yeah, but it doesn’t work for rings without one (like $2ℤ$) either. – k.stm Apr 11 '15 at 11:18

1) if $G$ is a finitely generated abelian group, then $G\simeq\mathrm{End}(G)$ iff $G$ is cyclic.
indeed, if $G$ is infinite and has an element of prime order $p$, then $\mathrm{End}(G)$ has more elements of order $p$ than $G$. A similar counting argument also works when $G$ is finite and non-cyclic. And if $G$ is free abelian of rank $r\ge 2$, then $\mathrm{End}(G)$ is free abelian of rank $r^2$.
2) if $G$ is a subring of $\mathbf{Q}$ (e.g., $\mathbf{Z}[1/n]$ or $\mathbf{Q}$ itself), then $G\simeq\mathrm{End}(G)$. On the other hand, there are subgroups of $\mathbf{Q}$ such that $G$ is not isomorphic to $\mathrm{End}(G)$: for instance the subgroup of $\mathbf{Q}$ generated by the $1/p$ for prime $p$ has $\mathrm{End}(G)\simeq\mathbf{Z}$.
3) there are some more examples: for instance if $A$ is a subring of $\mathbf{Q}$ containing $1/n$ then $G=A\times(\mathbf{Z}/n\mathbf{Z})$ satisfies $G\simeq\mathrm{End}(G)$.
4) if $G\simeq H^{(X)}$ for some nonzero countable group $H$ and infinite set $X$ (this is the set of finitely supported functions $X\to H$), then $G$ is not isomorphic to $\mathrm{End}(G)$, just because $\mathrm{End}(G)$ has cardinal greater than $G$: indeed the cardinal of $G$ is that of $X$, while the cardinal of $\mathrm{End}(G)$ is that of $2^X$ (because $\mathrm{End}(G)$ contains $\mathrm{End}(H)^X$).