# Image of a map $\mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z}$

I am breaking off a discussion with Mariano from here

I have a map $\mathbb{Z} \stackrel{\left(\begin{array}{c}2\\2\end{array}\right)}{\to} \mathbb{Z} \oplus \mathbb{Z}$

This is the map $(a) \mapsto (2a,2a)$. I thought that the image of this map was just $2 \mathbb{Z} \oplus 2\mathbb{Z}$. But to quote Mariano:

"the image of a map $\mathbb{Z}\to \mathbb{Z}\oplus \mathbb{Z}$ is either zero or a subgroup of rank 1, and $2\mathbb{Z}\oplus 2\mathbb{Z}$ has rank two."

I am trying to understand this further (I guess both Mariano's comment, and what the image would actually be)

-

$2\mathbb{Z}\oplus2\mathbb{Z}$ comprises all pairs of the form $(2a,2b)$, whereas you only have pairs of the form $(2a,2a)$. The image of your map is in fact isomorphic to $2\mathbb{Z}$; the doubling is redundant.

-
It is that simple! Not a great question then. It is Friday I guess... – Juan S Mar 25 '11 at 6:30

The image, of course, is the set of all pairs $(2a, 2a)$ where $a$ is an integer - in other words, it is the set of all pairs $(m, m)$ where $m$ is an even integer. If you visualize $\mathbb{Z} \oplus \mathbb{Z}$ as a grid, this is simply the diagonal with every other point removed.

The group $2\mathbb{Z} \oplus 2\mathbb{Z}$ is quite different: You have an even number in the first position, and an even number in the second position, but nothing says those two numbers are the same. So it's the set of pairs $(m,n)$ where both $m$ and $n$ are even. Plot this on the same grid and you'll see the difference.

-

(There was no need to ask a new question IMO... Oh well)

The group $\mathbb Z$ is cyclic, generated by $1$. It follows that the image of any group homomorphism $\phi:\mathbb Z\to G$ is generated as a subgroup of $G$ by $\phi(1)$ and, in particular, it is cyclic.

Now if $G$ is torsion free, then $\phi$ or injective or zero. Otherwise, $\phi$ would induce an injective map $\mathbb Z/\ker\phi\to G$ and this is impossible because $\mathbb Z/\ker\phi$ would then be a non-zero torsion group.

We conclude that if $G$ is torsion-free, the image of a map $\phi:\mathbb Z\to G$ is either $0$ or isomorphic to $\mathbb Z$.

-
@Marino - sorry - it was easier than I was thinking (again, I'll blame Friday) – Juan S Mar 25 '11 at 6:33