Let $G$ and $H$ be abelian groups and let $\varphi:G\rightarrow H$ be a group homomorphism. Can we define a nontrivial scalar multiplication on $G$ (from some field $F$ ) and another nontrivial scalar multiplication on $H$ (from $F$ ), thereby making them vector spaces, such that $\varphi$ becomes a linear map? If the answer is yes, then would this also work if we made $G$ and $H$ into modules over some ring $R$?


4 Answers 4


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 $g=0$.

  • 1
    $\begingroup$ Your last paragraph is incomplete - all you've shown is that no finite group can be the underlying group of a vector space over a field of characteristic $0$. But there are fields of nonzero characteristic! Andreas' answer below gives the right approach to building a finite group which cannot be a vector space. $\endgroup$ Commented Oct 28, 2016 at 17:05
  • $\begingroup$ @NoahSchweber: I agree with you: my last paragraph was meant only to show that in general the problem has a negative answer. Your argument is much better, since it shows that if the characteristic of the field is $\ne 2$, the only possible multiplication by scalars is the trivial one: $q \cdot g = 0$. $\endgroup$
    – Alex M.
    Commented Oct 28, 2016 at 17:18
  • 1
    $\begingroup$ @AlexM. The "trivial multiplication" isn't a valid one - vector spaces satisfy $1_k\cdot v=v$. And my argument does not only apply to characteristic $\not=2$ - note that I first show the characteristic must be $0$, so $\mathbb{Z}$ cannot be the underlying group of a vector space over any field. $\endgroup$ Commented Oct 28, 2016 at 17:22

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.


Not ANY morphism pf groups allows this, however I will show a sufficient condition for this to be true.

In the following question

Prove this homomorphism gives an equivalent definition of Linear Space

we provide an equivalent definition to linear space, in terms of a homomorphism from the product group of a field, and the group of automorphisms of an abelian group. There we prove how this homomorphism provides us with a definition of scalar prodcut, and the elements of the vector space are obviously the elements of the abelian group.

Let $G$ and $H$ be turned into vector spaces by this manner, and using the same field. This means we have two homomorphisms $*_1:\mathbb K(\cdot)\rightarrow\{GfG\}$ and $*_2:\mathbb K(\cdot)\rightarrow\{HfH\}$, here we use $\{GfG\}$ to represent the group of automorphisms of $G$. We know $*_1(a)$, for any $a\in\mathbb K$, is an automorphism of $G$, and this defines our scalar product, the same is true for $*_2(a)$.

We know $\phi(u+v)=\phi u+\phi v$. Then, we can say that a functor (homomorphism) $\phi:G\rightarrow H$ is a linear map if the following condition holds:

$$\phi\circ*_1(a)=*_2(a)\circ\phi$$ for every $a$ in the field. This condition simply means $\phi(au)=a(\phi u)$, for every element in th field.

In conclusion, if $G$ and $H$ can be made into linear spaces by means of the homomorphisms $*_1:G\rightarrow\{GfG\}$ and $*_2:H\rightarrow\{HfH\}$, and if the morphism $\phi$ commutes with the morhpsims in $Im(*_1)$, $Im(*_2)$, in the sense we just made clear, then we can say $\phi$ is a linear transformation from the linear space $G$ into the linear space $H$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .