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Let $M,N$ be $R$-modules; $A \subset M, B \subset N$ be their respective submodules. Show$\frac {M\oplus N}{ A\oplus B} \simeq \frac M A\oplus \frac N B$.

This is a book excercise I was working on, below is the proof I have been working on; I wanted to know if I made any mistakes.

Let $A'$ be a submodule where $A \oplus A'=M$ and $B'$ be a submodule where $B \oplus B'=N$.

LHS $\simeq \frac {A \oplus A' \oplus B \oplus B'}{A\oplus B} \simeq A' \oplus B'$

$\frac M A \simeq \frac {A \oplus A'}A \simeq A'$ and $\frac N B \simeq \frac {B \oplus B'}B \simeq B' \Rightarrow $ RHS $\simeq A' \oplus B'$

LHS=RHS

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    Just a silly comment: undergraduate students would love to have $$\frac{a+b}{c+d}=\frac{a}{c}+\frac{b}{d}$$ for positive numbers $a,b,c,d$, too. – 2017-01-24
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    @JackD'Aurizio This comment kind of suggests that the questioner’s statement is false as well. – 2017-01-24

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Your argument is wrong. In general you cannot find $A'$ such that $A\oplus A'=M$.

An example: $R=\mathbb{Z}$, $M=\mathbb{Z}$ and $A=2\mathbb{Z}$.

Consider the homomorphism $$ M\oplus N\to\frac{M}{A}\oplus\frac{N}{B}, \qquad (x,y)\mapsto (x+A,y+B) $$ What's the kernel? Is the homomorphism surjective?

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    So it would be a surjective since $\frac M A \subseteq M$ and $\frac N B \subseteq N$ ? And then would the kernal be $A \oplus B$? – 2017-01-25
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    @Mark That's the idea – 2017-01-25
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In the denominator of the left hand side – that is, $A \oplus B$ in $\frac{M \oplus N}{A \oplus B}$ – the symbol “$\oplus$” refers to the internal direct sum.

There need not be a direct summand $A'$ for $A$ adding up to $M = A \oplus A'$ internally (and not even externally) – the classic example: Let $M = ℤ$ and $A = 2ℤ$ (and $N = 0$ and $B = 0$). There is no $ℤ$-submodule $A'$ of $M$ with $M = A \oplus A'$ internally. (And for $M = ℤ/4ℤ$ and $A = 2ℤ/4ℤ$, there even is no $ℤ$-module $A'$ with $M = A \oplus A'$ externally.)

Instead you need to employ the first isomorphism theorem for a suitable module homomorphism $M \oplus N → \frac M A \oplus \frac N B$.