if $K=L$ then $K'\cong L'$. Furthermore, if $H$ is submodule of $M$ and $K$ is submodule of $H$ then $H=K\oplus\left(H\cap K'\right)$

Question

Let $M$ be $R$-module and $K,K',L,L'$ be submodules of $M$ with the property that $M=K\oplus K'=L\oplus L'$. Prove that if $K=L$ then $K'\cong L'$. Furthermore, if $H$ is submodule of $M$ and $K$ is submodule of $H$ then $H=K\oplus\left(H\cap K'\right)$

Thanks for any insight.

Answer 1

First thing: suppose $K\oplus K'=K\oplus L'$ then $K'\cong (K\oplus K')/K=(K\oplus L')/K\cong L$.

Second thing: by distributivity you get $H=(K+ K')\cap H=(K\cap H)+ (K'\cap H)=K+(K'\cap H)$. Notice that $K\cap (K'\cap H)\subset K\cap K'=0$ so $H=K\oplus(K'\cap H)$.