When is a non-trivial homomorphism injective?

Question

I noticed that over the natural numers $(\mathbb{Z},+)$ any group homomorphism $f : \mathbb{Z} \rightarrow \mathbb{Z}$ that is not the trivial one, is automatically injective.

Where exactly does this come from?- Thus, I want to ask: What is the abstract reason for this behaviour?

Answer 1

The reason is because $\mathbb Z$ is cyclic and infinite - it is generated by $1$. Hence, the image of a homomorphism $\theta:\mathbb Z \to \mathbb Z$ will be generated by $\theta(1)$.

If $\theta(1) = 0$ this gives the trivial homomorphism. Otherwise, $\theta(1) = a \in \mathbb Z$ and therefore $\theta(n) = an \quad \forall n \in \mathbb Z$.

In general, if $C$ is an infinite cyclic group and $\phi:C \to C$ is a homomorphism, then $\phi$ is either trivial or injective.

However, if $C_n$ is cyclic of order $n$ where $n$ is composite, there will be non-injective non-trivial homomorphisms. For example, if $p \mid n$, then $\theta: x \mapsto x^p$ is a non-injective homomorphism.

Answer 2

One possible answer is that any such endomorphism becomes multiplication by an element in the ring $Z$ (the ring of endomorphisms of Z), and so lifts to an endomorphism on the field of fractions (the rationals), where it is seen to be injective - in general field homomorphisms are injective because fields have no nontrivial ideals.

P.S. The natural numbers are the nonnegative integers.

But the integers are kind of special in this regard. The endomorphism ring of a group is almost never going to be a domain (for instance, if the group can be written as a product), so it won't be possible to form a field of fractions and repeat this argument. So maybe an abstract reason is because the ring of endomorphisms of Z is a domain?

Answer 3

One issue is torsion.

If $f:\mathbb{Z}\to A$ is a homomorphism of abelian groups, we have $\ker f = n\mathbb{Z}$ for some $n$. If $n>1$, then $f(\mathbb{Z})$ has $n$-torsion, namely $n\cdot f(1) = f(n) = f(0)$.

So if $A$ is a group with no torsion (say, a free abelian group), the only morphisms $f:\mathbb{Z}\to A$ are either trivial ($n=1$) or injective ($n=0$).

To connect this to user54092's answer: if we consider a morphism of free abelian groups $f:B\to A$, we can always lift to a morphism of $\mathbb{Q}$-vector spaces $f\otimes {id}_\mathbb{Q} : B\otimes \mathbb{Q} \to A\otimes \mathbb{Q}$. So we can make use of similar theorems for vector spaces, namely that there are no non-trivial subspaces of $\mathbb{Q}^1$.