I'm reading Rotman- Advanced Modern Algebra p.566, and I got stuck in the proof for Lemma 8.61-(iii)
Here are some lemmas the author introduced to prove lemma 8.61: (Here, all rings are assumed to have a unity)
Lemma 8.53 -(i)
Let $R$ be a ring and $I,J$ be minimal left-ideals of $R$. If $I^2\neq 0$ and $I\cong J$ as $R$-modules, then $IJ\neq 0$.
Lemma 8.53 -(ii)
Let $R$ be a ring and $I,J$ be minimal left-ideals of $R$ such that $IJ\neq 0$. Then, $Hom_R(I,J)\neq 0$ and there exists $b'\in J$ such thay $J=Ib'$.
Now here is a part of the proof for Lemma8.61-(iii) where I got stuck:
Let $R$ be a semisimple ring and $B_1,...,B_n$ be the simple components of $R$. Let $D$ be a nonzero two-sided ideal in $R$. Since $R$ is semisimple, $R$ is Artinian. Thus, there exists a minimal left-ideal $L$ of $R$ such that $L\subset D$. Since $L$ is minimal, $L\subset B_i$ for some $i$. We claim that $B_i\subset D$. Let $L'$ be another minimal-left ideal of $R$ such that $L'\subset B_i$. (Note that $L\cong L'$ as $R$-modules in this case) Then, there exists $b'\in L'$ such that $L'=Lb'$.
I cannot figure out how this last sentence holds true. Using lemma 8.53-(i),(ii), it suffices to prove that $L^2\neq 0$. But how?