= Solution
An <R-module> $M$ is a <free module> if it has a basis $(e_i)_{i\in I}$: every $m\in M$ has a unique expression as a finite sum $\sum_i r_ie_i$, with $r_i\in R$. Equivalently, it is a <direct sum> of copies of $R$. Let $q:M\to M/N$ be the quotient map. When $M/N$ is free, choose a basis $(\overline f_j)$ and lifts $f_j\in M$. The <universal property of a free module> defines an $R$-linear map $s:M/N\to M$ by $s(\sum_jr_j\overline f_j)=\sum_jr_jf_j$. It satisfies $q\circ s=\operatorname{id}$, so every $m\in M$ decomposes uniquely as
$$
m=\bigl(m-s(q(m))\bigr)+s(q(m)),\qquad m-s(q(m))\in N.
$$
This gives a <split short exact sequence> and $M\cong N\oplus(M/N)$. If $N$ is also free, the union of its basis and the lifted quotient basis is a basis of $M$: spanning follows from the decomposition, and applying $q$ to a putative relation kills all quotient coefficients before independence in $N$ kills the remaining ones. Thus \b[free submodule and free quotient imply a free module].
For the first claim, take $R=\mathbb Z$ and
$$
M=\mathbb Z,\quad N=2\mathbb Z,\qquad M'=\mathbb Z\oplus\mathbb Z/2\mathbb Z,\quad N'=\mathbb Z\oplus\{0\}.
$$
Here $N\cong N'\cong\mathbb Z$ are <free modules> and $M/N\cong M'/N'\cong\mathbb Z/2\mathbb Z$. But $M$ is a <torsion-free module>, whereas $M'$ contains the nonzero <torsion element> $(0,\overline1)$. Therefore \b[claim (1) is false].
For the second claim, the same lifting construction works without assuming that $N$ is free. Since both isomorphic <quotient modules> are free, it yields
$$
M\cong N\oplus(M/N)\cong N'\oplus(M'/N')\cong M'.
$$
Thus \b[claim (2) is true]. Equivalently, this uses the fact that <free modules are projective>.
Back to article page
