Following L. Fuchs’ book Infinite Abelian Groups, we give in this note three equivalent definitions of divisibility of a group.

Definition: A divisible group $G$ is an abelian group such that for any $x \in G$ and $n \geq 1$ there exists $y \in G$ satisfying $x=ny$.

Definition: An injective group $D$ is an abelian group such that for any abelian groups $A$ and $B$ and any morphisms $j : A \hookrightarrow B$ and $\varphi : A \to D$ there exists a morphism $\eta : B \to D$ so that $\eta \circ j=\varphi$.

Theorem 1: A group is injective if, and only if, it is divisible.

Proof. Let $D$ be a divisible group. For $x \in D$ and $n \geq 1$, let $\eta$ be the morphism given by the definition where $A= \langle x \rangle$ and $B$ is a divisible group containing $A$ (eg. $A= \mathbb{Q}$ if $x$ has infinite order or $A$ is a direct sum of Prüfer groups otherwise).

Because $B$ is divisible, there exists $y \in B$ such that $j(x)=ny$, hence $x= \eta \circ j(x)=n \cdot \eta(y)$. Therefore, $D$ is divisible.

Let $D$ be a divisible group. Let $A,B$ be two abelian groups and $j : A \hookrightarrow B$, $\varphi : A \to D$ be two morphisms. Without loss of generality, we suppose that $A$ is a subgroup of $B$ and that $j$ is just the canonical injection.

Let $\mathcal{P}$ be the set of pairs $(C,\theta)$ where $A \leq C \leq B$ and $\theta : C \to D$ extends $\varphi$, ordered by:

$(C_1,\theta_1) \prec (C_2,\theta_2)$ if $C_1 \leq C_2$ and $\theta_2$ extends $\theta_1$.

Notice that $(\mathcal{P},\prec)$ is inductive since every chain $\{(C_i,\theta_i) \mid i \in I\}$ is bounded above by $\left( \bigcup\limits_{i \in I} C_i , \bigcup\limits_{i \in I} \theta_i \right) \in \mathcal{P}$. According to Zorn’s lemma, there exists a maximal element $(C,\theta)$ of $\mathcal{P}$.

By contradiction, suppose that $C \subsetneq B$ and let $b \in B \backslash C$.

• Case 1: $C \cap \langle b \rangle= \langle nb \rangle$ where $n \geq 1$.

Because $D$ is divisible there exists $d \in D$ such that $\theta(nb)=nd$. Introduce:

$\psi : \left\{ \begin{array}{ccc} \langle C,b \rangle & \to & D \\ c+mb & \mapsto & \theta(c)+md \end{array} \right.$ where $c \in C$ and $0 \leq m < n$.

First, $\psi$ is well-defined since any element of $\langle C,b \rangle$ can be uniquely written as $c+mb$ where $c \in C$ and $0 \leq m . Indeed, suppose that $c+mb=c'+m'b$ where $c,c' \in C$ and $0 \leq m,m' < n$. Then $(m-m')b=c'-c \in C \cap \langle b \rangle =\langle nb \rangle$ hence $m=m'$ and $c=c'$ since $-n.

Secondly, $\psi$ is a morphism: if $c,c' \in C$ and $0 \leq m,m' , writing the division $m-m'=kn+p$, we have:

$\begin{array}{ll} \psi((c+mb)-(c'+m'b)) & = \psi((c-c'+knb)+pb) = \theta(c-c'+knb)+pd \\ \\ & = \theta(c)-\theta(c') +k \underset{nd}{\underbrace{\theta(nb)}}+pd \\ \\ & = \theta(c)-\theta(c')+ \underset{m-m'}{\underbrace{(kn+p)}}d \\ \\ & = ( \theta(c)+md)-(\theta(c')+m'd) \\ \\ & = \psi(c+mb)-\psi(c'+m'b) \end{array}$

• Case 2: $C \cap \langle b \rangle = \{0\}$.

Taking any $d \in B$, we define in a similar way:

$\psi : \left\{ \begin{array}{ccc} \langle C,b \rangle & \to & D \\ c+mb & \mapsto & \theta(c)+md \end{array} \right.$ where $c \in C$ and $m \in \mathbb{Z}$.

Again, $\psi$ is a well-defined homomorphism.

Finally, $(\langle C,b \rangle,\psi) \in \mathcal{P}$ contradicts the maximality of $(C,\theta)$. Consequently, $C=B$ and it is sufficient to set $\eta:= \theta$ to conclude that $D$ is injective. $\square$

In an abelian group $A$, a system of equations can be written as

$\displaystyle \sum\limits_{j \in J} n_{ij}x_j=a_i$, $i \in I$

where $a_i \in A$, $n_{ij} \in \mathbb{Z}$ and where $\{ n_{ij} \neq 0 \mid j \in J\}$ is finite for each $i \in I$. Formally, $f_i := \sum\limits_{j \in J} n_{ij}x_j$ can be viewed as an element of the abelian free group $F$ over $\{x_j \mid j \in J\}$.

We say that the system of equations is compatible over $A$ if a linear combination of $f_i$‘s leads to a linear combination of $a_i$‘s, ie. if $f_k= \sum\limits_{m \in M} f_m$ then $a_k= \sum\limits_{m \in M} a_m$ (where $M$ is finite). Clearly, if the system has a solution over $A$ then it is necessarily compatible.

Theorem 2: Let $A$ be an abelian group. Every compatible system of equations over $A$ has a solution in $A$ if, and only if, $A$ is a divisible group.

Proof. First, notice that the following system of equations is compatible over $A$:

$nx_{a,n} =a$ with $n \geq 1$ and $a \in A$

Therefore, if every compatible system of equations has a solution, then $A$ is a divisible group.

Conversely suppose that $A$ is a divisible group and consider a compatible system of equations overs $A$:

$\displaystyle \sum\limits_{j \in J} n_{ij}x_j=a_i$ with $i \in I$ and $n_{ij} \in \mathbb{Z}$.

As above, let $X$ be the free abelian group over $\{ x_j \mid j \in J\}$ and $Y$ be the subgroup generated by $\{ f_i:= \sum\limits_{j \in J} n_{ij}x_j \mid i \in I\}$.

Because the system is compatible, $f_i \mapsto a_i$ extends to a homomorphism $\chi : Y \to A$. According to theorem 1, $A$ is an injective group, so $\chi$ extends to a homomorphism $\eta : X \to A$. Finally, $\{ \eta(x_i) \mid i \in I\}$ is a solution to the system. $\square$

Corollary: A system of equations over a divisible group $A$ has a solution if, and only if, every finite subsystem has a solution.

Just for fun, let us mention the following property:

Property: A subgroup $B$ of an abelian group $A$ is a direct summand if, and only if, every system of equations over $B$ solvable in $A$ is solvable in $B$.

Proof. If $B$ is a direct summand of $A$, then the projection on $B$ of any solution of a system of equations over $B$ is a solution in $B$.

Conversely, suppose that every system of equations over $B$ solvable in $A$ is solvable in $B$. For any coset $u \in A/B$ let $a(u) \in A$ be a representant and consider the following compatible system of equations over $B$:

$x_u+x_v-x_{u+v} = a(u)+a(v)-a(u+v) \in B$ with $u,v \in A/B$.

By assumption, there exists a solution $\{b(u) \mid u \in A/B\}$. Let $C= \{ a(u)-b(u) \mid u \in A/B\}$; by construction, $C$ is a subgroup of $A$. Moreover, $A=B+C$ and $B \cap C = \{0\}$. Therefore, $B$ is a direct summand. $\square$

Corollary: Any divisible subgroup of an abelian group is a direct summand.