Let $R$ be a ring with unity and $M$ any right $R$-module. Recall that a submodule $B\subseteq M$ is called essential if $B\cap L \neq 0$ for every nonzero submodule $L\subseteq M$. In this case, we write $B \subseteq^{ess} M$.
Recall that $M$ is called CS if every submodule of $M$ is essential in a summand (i.e., if $A\subseteq M$ is a submodule, there exists $L\subseteq M$ with $A \subseteq^{ess} L$).
Recall that $M$ is called summand-square-free if whenever $A,B$ are summands of $M$ with $A\cong B$ and $A \cap B=0$ then $A=B=0$. the module $M$ is called square-free if whenever $A,B$ are submodules of $M$ with $A\cong B$ and $A\cap B=0$ then $A=B=0$.
I need to prove that any CS summand-square-free module is square-free.
My attempt: Assume that $A,B\subseteq M$ are submodules with $A\cong B$ and $A\cap B=0$. Since $M$ is CS, there exists summands $X,Y\subseteq M$ with $A \subseteq^{ess} X$ and $B \subseteq^{ess} M$. Clearly, $X \cap Y=0$. If $X \cong Y$, we are done.
How can I proceed ?!.