**Proof plan.** The underlying additive group structure is already handled by the [First Isomorphism Theorem for Groups](/theorems/842). We additionally verify that the induced map respects the scalar action. **Step 1: $\ker(f)$ is a submodule of $M$.** [claim: Kernel Is Submodule] $\ker(f) = \{m \in M : f(m) = 0\}$ is an $R$-submodule of $M$. [/claim] [proof] $\ker(f)$ is a subgroup of $(M, +)$ since $f$ is a [group](/page/Group) homomorphism. For scalar closure: if $m \in \ker(f)$ and $r \in R$, then $f(r \cdot m) = r \cdot f(m) = r \cdot 0 = 0$, so $r \cdot m \in \ker(f)$. [/proof] **Step 2: Define the quotient map.** By the [First Isomorphism Theorem for Groups](/theorems/842), the additive group map \begin{align*} \bar{f} : M/\ker(f) &\to \operatorname{im}(f) \\ m + \ker(f) &\mapsto f(m) \end{align*} is a well-defined bijective group homomorphism. **Step 3: $\bar{f}$ respects the scalar action.** [claim: Scalar Compatibility] $\bar{f}$ is an $R$-module homomorphism. [/claim] [proof] For any $r \in R$ and coset $m + \ker(f)$: \begin{align*} \bar{f}(r \cdot (m + \ker f)) = \bar{f}(r \cdot m + \ker f) = f(r \cdot m) = r \cdot f(m) = r \cdot \bar{f}(m + \ker f). \end{align*} [/proof] Since $\bar{f}$ is a bijective $R$-module homomorphism, $M/\ker(f) \cong \operatorname{im}(f)$. $\square$