[proofplan] We compare the defining closure conditions. A $\mathbb{Z}$-submodule is already an additive subgroup of the underlying abelian group. Conversely, if $B$ is an additive subgroup, then it is closed under all integer scalar multiples because positive multiples are repeated sums, the zero multiple is the identity element, and negative multiples are additive inverses of positive multiples. [/proofplan] [step:Show that every $\mathbb{Z}$-submodule is an additive subgroup] Assume that $B$ is a $\mathbb{Z}$-submodule of $A$. By the definition of submodule, $B$ is an additive subgroup of the underlying additive group of $A$: it contains $0_A$, and for all $x,y \in B$, both $x+y \in B$ and $-x \in B$. Therefore $B$ is a subgroup of $A$. [/step] [step:Show that every additive subgroup is closed under integer scalar multiplication] Assume conversely that $B$ is a subgroup of the additive group $A$. Then $0_A \in B$, and for all $x,y \in B$, one has $x+y \in B$ and $-x \in B$. Let $b \in B$ and let $n \in \mathbb{Z}$. We prove that $nb \in B$. If $n=0$, then $nb=0_A \in B$. If $n \in \mathbb{N}$, then $nb$ is the sum of $n$ copies of $b$. Since $B$ is closed under addition, induction on $n$ gives $nb \in B$. If $n<0$, define $m=-n \in \mathbb{N}$. The positive case gives $mb \in B$, and closure under additive inverses gives \begin{align*} nb = -mb \in B. \end{align*} Thus $B$ is closed under scalar multiplication by every integer. [guided] We must verify the only submodule condition that is not already part of being an additive subgroup: closure under scalar multiplication by elements of $\mathbb{Z}$. Fix $b \in B$ and $n \in \mathbb{Z}$. There are three cases. If $n=0$, then the canonical $\mathbb{Z}$-module structure gives \begin{align*} nb = 0b = 0_A. \end{align*} Since $B$ is a subgroup of $A$, it contains the identity element $0_A$, so $nb \in B$. If $n \in \mathbb{N}$, then $nb$ means the repeated sum of $n$ copies of $b$. Because $b \in B$ and $B$ is closed under the addition operation inherited from $A$, a direct induction on $n$ shows that this repeated sum lies in $B$. The base case is $1b=b \in B$. For the induction step, if $kb \in B$ for some $k \in \mathbb{N}$, then \begin{align*} (k+1)b = kb + b \in B \end{align*} because both $kb$ and $b$ lie in $B$ and $B$ is closed under addition. Finally, suppose $n<0$. Define $m=-n$, so $m \in \mathbb{N}$. By the positive case, $mb \in B$. Since $B$ is a subgroup, it is closed under additive inverses, hence \begin{align*} nb = -mb \in B. \end{align*} Therefore $nb \in B$ for every $n \in \mathbb{Z}$ and every $b \in B$. [/guided] [/step] [step:Conclude that the two structures have the same subsets] From the subgroup assumption, $B$ contains $0_A$, is closed under addition, is closed under additive inverses, and, by the previous step, is closed under scalar multiplication by every integer. These are exactly the closure conditions for $B$ to be a $\mathbb{Z}$-submodule of $A$. Hence $B$ is a $\mathbb{Z}$-submodule of $A$ if and only if $B$ is a subgroup of the additive group $A$. [/step]