Isomorphism of quotient of direct sum modules Let $M, N, M'$ and $N'$ be R-modules. If $M'$ and $N'$ are submodules of both $M$ and $N$ then is it true that
\begin{equation}
\frac{M}{M'} \oplus \frac{N}{N'} \cong \frac{M \oplus N}{M' \oplus N'} \cong \frac{M \oplus N}{N' \oplus M'} \cong \frac{M}{N'} \oplus \frac{N}{M'} ?
\end{equation}
I think the first and the last equality is true and I could find an isomorphisms for them, but the middle equality is what confuses me. I feel that it should be true since $M' \oplus N' \cong N' \oplus M'$ but if I try to define an isomorphism as something like
\begin{equation}
\psi : (m + M', n + N') \rightarrow (m + N', n + M')
\end{equation}
there is a difficulty because $m_1 + M' = m_2 + M'$ doesn't really implies $m_1 + N' = m_2 + N'$. So at the end I'm not sure what a conclusion is. Could someone give an answer with explaination to resolve is paradox please, thank you.
 A: Your claim is NOT true.
$M = \mathbb{Z}, N = \mathbb{2Z}, M' = \mathbb{8Z}, N' = \mathbb{4Z}.$ All are considered modules over $\mathbb{Z}.$ Then $\frac{M}{M'} \oplus \frac{N}{N'} \cong \mathbb{Z}_8 \oplus \mathbb{Z}_2.$ On the other hand $\frac{M}{N'} \oplus \frac{N}{M'} \cong \mathbb{Z}_4 \oplus \mathbb{Z}_4.$ So this two are not isomorphic.
The mistake was to assume that since the submodules are isomorphic, so the quotients are isomorphic. This is not true in general. For example, take $M = \mathbb{Z}, M_1 = \mathbb{2Z}, M_2 = \mathbb{3Z}$ (as $\mathbb{Z}$ modules). Then $M_1 \cong M_2$, but $M/M_1 \ncong M/M_2.$ The main difficulty here is that the isomorphism between $M_1$ and $M_2$ is not induced by an isomorphism of $M$(why?). But if the isomorphism of $M_1$ and $M_2$ is induced by an isomorphism of $M$, then we will have $M/M_1 \cong M/M_2.$ Here is the actual statement: Let $A$ be a commutative ring with identity, $M$ is a $A$ module, and let $M_1, M_2$ be two submodules of $M.$ Suppose that $f: M_1 \rightarrow M_2$ is an $A$ module isomorphism. Assume that $f$ can be lifted to an $A$ module isomorphism $g$ of $M$, i.e. $g : M \rightarrow M$ is an $A$ module isomorphism, $g(M_1) = M_2$ and $g|M_1 = f.$ Then $M/M_1 \cong M/M_2.$ To see this, define a map $\phi : M \rightarrow M/M_2$ by $m \mapsto g(m) + M_2.$ Then show that $\phi$ is an $A$ module surjective homomorphism and Ker$\phi = M_1.$ (I think this is true over non-commutative ring also.)
