Find a countable abelian group whose endomorphism ring has the same cardinality as the set of real numbers. Justify your answer.
Solution
Let \(G\) be a vector space of dimension \(\aleph_0\) over \(\FF_2\). Then \(G\) is countable, since it is a countable union of finite subspaces. Let \(v_1,v_2,\ldots\) be a basis. For each \(S\subseteq\{1,2,3,\ldots\}\) there is an endomorphism of \(G\) mapping each \(v_i\) to \(v_i\) or \(0\) according to whether \(i\in S\). Different subsets \(S\) give different endomorphisms, so \(\#\operatorname{End}G\geq2^{\aleph_0}\). On the other hand,
\[ \#\operatorname{End}G\leq(\#G)^{\#G}=\aleph_0^{\aleph_0}\leq(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\aleph_0}=2^{\aleph_0}. \]
Thus \(\#\operatorname{End}G=2^{\aleph_0}=\#\RR\).