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I am trying to solve this exercise in the Book of Roman: if $S$ and $T$ are isomorphic submodules of a module $M$ it does not necessarily follow that $M/S\approx M/T$. Prove that this statement does hold if all modules are free and have finite rank.

I have a counterexample but not a proof.

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    Since you are already given that $M/S$ and $M/T$ are free, all you have to prove is that their ranks are the same. Now prove this statement: rank($S)+$rank$(M/S)=$rank$(M)$. – 2017-02-14
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    I am undecided if I also consider the quotients as free modules of finite range. If so, then it is very easy. – 2017-02-14
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    You have to, because otherwise the statement isn't true: Say $R=M = \mathbb{Z}$ and $S=M$ and $T=2M$. – 2017-02-14
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    I see. Thanks for all. – 2017-02-14
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    One of the two of you should post an answer to resolve this post. – 2017-02-15

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We have $\textrm{rank}\left(M\right)=\textrm{rank}\left(S\right)+\textrm{rank}\left(M/S\right)$ and $\textrm{rank}\left(M\right)=\textrm{rank}\left(T\right)+\textrm{rank}\left(M/T\right)$. Since $S\approx T$ then $\textrm{rank}\left(M/S\right)=\textrm{rank}\left(M/T\right)$ . Therefore $M/S\approx M/T$ .