Is it possible to make any abelian group homomorphism into a linear map?

Every abelian group is a $\mathbb{Z}$-module, so we can always treat $\varphi$ as a linear map of $\mathbb{Z}$-modules.

However, it is not always possible to view an abelian group as a vector space over a field. For example, the abelian group $\mathbb{Z}$ (with addition) is not a vector space over any field.

Why not? Well, suppose $\mathbb{Z}$ was (the underlying group of) a vector space over a field $k$. First, note that $k$ must have characteristic $0$ - if $k$ had characteristic $p\not=0$ then we'd have $1+1+...+1$ ($p$ times) equals $0$ in $\mathbb{Z}$.

So $k$ has characteristic $0$, which means $\mathbb{Q}$ is a subfield of $k$. (In fact, we may assume without loss of generality that $k=\mathbb{Q}$, but we don't need to.) What can ${1\over 2}\cdot 1$ be (where the ${1\over 2}$ is in $k$ and the $1$ is in $\mathbb{Z}$)? Whatever integer this is, it must satisfy ${1\over 2}\cdot 1+{1\over 2}\cdot 1=1$. But there is no integer with this property.


Notice that any commutative $G$ is a $\Bbb Z$-module in a natural way: just define $n \cdot g$ to be $g^n$. Check that the usual properties for modules are satisfied.

Now, if $\varphi$ is a group morphism, then $\varphi (g^n) = \varphi (g) ^n$, or in our $\Bbb Z$-module notation: $\varphi (n \cdot g) = n \cdot \varphi (g)$, i.e. $\varphi$ is $\Bbb Z$-linear.

On the other hand, not all groups can be given a vector space stucture. For instance, if $G$ is a finite group (write it in additive notation) and $d = |G|$, and if $K \supseteq \Bbb Q$, then $d \cdot g = 0$ for all $g \in G$. Multiplying by the scalar $\frac 1 d \in \Bbb Q$ would give $g=0$.


To record another example, a finite vector space has order a power of a prime. So if the order of the abelian group $G$ is not a power of a prime, it cannot be turned into a vector space.

Actually, even if the abelian group $G$ is of order $p^{n}$, for some prime $p$, then if it can be turned into a vector space over a field $F$, then it will also be a vector space over the field with $p$ elements. And then $p x = 0$ has to hold for all $x \in G$. Conversely, if this conditions holds, then even if $G$ is not finite, it can be turned into a vector space over the field with $p$ elements.


Addendum. Even if $G$ and $H$ can be turned into vector spaces, it is not always the case that $\phi$ can be made linear. In fact, if $G$ and $H$ can be made into vector spaces over a field $F$, then they can be made into vector spaces over the prime field $P$ contained in $F$, thus over the rationals if $F$ has characteristic zero, over the field with $p$ elements if $F$ has prime characteristic $p$. And any group homomorphism from $G$ to $H$ will be a $P$-linear map.

But take for instance $G = H = \mathbb{R}$. Then the only $\mathbb{R}$-linear maps from $G$ to $H$ are the multiplications by a fixed real number. Whereas there are many more group homomorphisms from $G$ to $H$, see Hamel bases. So if you make $G$ and $H$ into rational vector spaces, all group homomorphisms will be linear. If you attempt to make them into real vector space, this is not the case anymore.