For groups, rings, or $R$-modules $A, B, C$ with $C\subseteq B$, does $A/B \cong A/C$ imply that $B=C$?

Question

I think this is true if the module is finite, using Lagrange's Theorem. However, is this the case for infinite modules (or rings, or groups)? This comes from me trying to prove that tensoring is right-exact; at some point in my proof, I am trying to show that if $0\to M\xrightarrow[]{f} N\xrightarrow[]{g} P\to 0$ is exact, and $\text{im}(f\otimes \text{id}_C) \subseteq \text{ker}(g\otimes \text{id}_C)$, then $(N\otimes C)/\text{im}(f\otimes \text{id}_C) \cong (N\otimes C)/\text{ker}(g\otimes \text{id}_C)$ implies that $\text{im}(f\otimes \text{id}_C)= \text{ker}(g\otimes \text{id}_C)$.

Answer 1

You can fit those modules into the following diagram,

So now your want to construct a morphism from $B$ to $C$, which seems is doable by diagram-chasing (though, I am lazy and did not check this, I will just let other people, or you, tell me if my guess is correct).