Assume that $M$ is any right $R$-module. Assume that $M$ has no infinite direct sums of simple summands. Does this assumption imply that $M$ has no infinite set of pairwise isomorphic simple summands ?!.
Here is my attempt: Assume on the contrary that $M$ has an infinite set $\lbrace S_i \rbrace_{i\in I}$ of pairwise isomorphic simple summands. There exists $\Omega \subseteq I$ maximal with respect to $\sum_{i\in \Omega}S_i$ is an internal direct sum. By the hypothesis that $M$ has no infinite direct sums of simple summands, $\Omega$ must be finite.
I couldn't complete the proof. Well, Can this argument lead to a proof ?!.
I appreciate your help. Thanks in advance.
$\mathbb R^2$ as an $\mathbb R$ module does not have "infinite direct sums of simple summands" (it contains a maximum of two in a direct sum) but it has an infinite set of distinct one-dimensional subspaces, all of which are pairwise isomorphic, and all of which are summands.