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Is it true that a group homomorphism $f:\mathbb Z\oplus \mathbb Z\to \mathbb Z\oplus \mathbb Z$ that is injective must be an isomorphism?

I know that non zero homomorphisms $g:\mathbb Z\to \mathbb Z$ are all injective but are isomorphisms only if $g(1)=\pm 1$. But the situation for homomorphisms of $f:\mathbb Z\oplus \mathbb Z\to \mathbb Z\oplus \mathbb Z$ is not clear to me.. I know that $f $ is completely determined by $f(1,0)$ and $f(0,1)$

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if $G_1$ and $G_2$ are two copies of $\mathbb Z\oplus \mathbb Z$ and $f:G_1\to G_2$ injective and we know that there exists $h:G_2\to G_1$ surjective (does this imply that $f$ is an isomorphism? –  palio Nov 30 '11 at 19:28
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Since a subgroup of $\mathbb{Z}^k$ is necessarily isomorphic to $\mathbb{Z}^{\ell}$ for some $\ell$, $0\leq \ell\leq k$, the image of any homomorphism $f\colon\mathbb{Z}^m\to\mathbb{Z}^n$ is necessarily isomorphic to $\mathbb{Z}^r$ for some $r$, $0\leq r\leq \min(m,n)$. Since the map $\mathbb{Z}^n\to\mathbb{Z}^n$ given by $(a,b)\mapsto (ka,kb)$, $k\neq 0$, is one-to-one but not onto for $|k|\gt 1$, no such map is necessarily onto. –  Arturo Magidin Nov 30 '11 at 19:31
    
Of course not. $h$ always exists (take the identity); you are placing absolutely no restrictions on $f$ by positing the existence of $h$. Now, if $h$ is a retraction (i.e., $h\circ f = \mathrm{id}_{G_1}$), then that means that $G_2\cong \mathrm{Im}(f)\oplus\mathrm{ker}(h)$, and that will mean that $f$ is an isomorphism. –  Arturo Magidin Nov 30 '11 at 19:33
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Perhaps you should have added all hypotheses, instead of omitting them? Please add the hypothesis to the question, but as an Addition so that it does not invalidate previous answers. –  Arturo Magidin Nov 30 '11 at 20:05
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Since the question does not reflect the accepted answer, and you have failed to include all relevant information into the question, I am downvoting the question as not a good question. If you edit the question to include all relevant information, I will remove the downvote. –  Arturo Magidin Nov 30 '11 at 21:25

4 Answers 4

up vote 3 down vote accepted

From the comments, it seems that the actual situation is:

Suppose that $f,h\colon\mathbb{Z}\oplus\mathbb{Z}\to\mathbb{Z}\oplus\mathbb{Z}$ are group homomorphisms, and that $h$ is a retraction of $f$; that is, $h\circ f = \mathrm{id}$. Is $f$ an isomorphism?

The answer in that situation is "yes".

Since $h\circ f$ is bijective, $f$ is one-to-one and $h$ is onto. It is also straightforward to verify that $$\mathbb{Z}\oplus\mathbb{Z}\cong \mathrm{Im}(f) \oplus \mathrm{ker}(h).$$ Both $\mathrm{Im}(f)$ and $\mathrm{ker}(h)$ are free abelian. Since $f$ is one-to-one, $\mathrm{Im}(f)$ is free abelian of rank $2$. Since the rank of a free abelian group is uniquely determined, this means that $$2 = \mathrm{rank}(\mathbb{Z}\oplus\mathbb{Z}) = \mathrm{rank}(\mathrm{Im}(f))+\mathrm{rank}(\mathrm{ker}(h)) = 2 + \mathrm{ker}(h),$$ so $\mathrm{ker}(h)$ is trivial. Therefore, $h$ is one-to-one and onto, hence bijective, and since $h\circ f = \mathrm{id}$, it follows that $f$ is the inverse of $h$, and an isomorphism.

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A more interesting exercise is what condition on $f(1,0) = (a,b)$ and $f(0,1) = (c,d)$ is needed for the homomorphism $f: {\mathbb Z} \oplus {\mathbb Z} \to {\mathbb Z} \oplus {\mathbb Z}$ to be an isomorphism? The answer is that $ad - bc = \pm 1$ is necessary and sufficient. Now generalize to homomorphisms of ${\mathbb Z}^n$ to itself.

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I believe $f(a,b)=(2a,2b)$ is an easy counterexample.

It is easy to see that this map is injective group homomorphism, but it is not surjective.


As Arturo Magidin pointed out in his comment, for any injective homomorphism $f:\mathbb Z\oplus \mathbb Z\to \mathbb Z\oplus \mathbb Z$ the image $\operatorname{Im} f$ is isomorphic to $\mathbb Z\oplus \mathbb Z$, i.e., it is a free Abelian group of rank 2. Even more, for any homomorphism $f:\mathbb Z\oplus \mathbb Z\to \mathbb Z\oplus \mathbb Z$ the image is generated by $f(1,0)$ and $f(0,1)$, which implies that it will be a free Abelian group. (It is explained more detailed in the comments.)

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Of course, the observation is a bit silly stated that way: any injective group homomophism gives an isomorphism onto its image. What I was thinking is "the image of any morphism must be trivial, cyclic, or free abelian of rank $2$". –  Arturo Magidin Nov 30 '11 at 19:28
    
@Arturo Sorry, it seems that I misunderstood your comments. Hopefully added text together with your comments clarifies the situation.) –  Martin Sleziak Nov 30 '11 at 19:36
    
You didn't misunderstand them: I expressed myself poorly (then deleted and rephrased). So no worries. –  Arturo Magidin Nov 30 '11 at 19:37

To follow up on Robert Israel's answer, consider the $n$-fold direct sum $R^{n}$ where $R$ is a commutative ring with $1$. A homomorphism $f:R^{n} \rightarrow R^{n}$ corresponds to an $n \times n$ matrix $M$ with entries in $R$. It turns out that $f$ is injective if and only if $\det (M)$ is a regular element (i.e., non zero divisor) in $R$ and an isomorphism if and only if $\det (M)$ is a unit of $R$, whence Robert's condition. So, every injective $f$ is an isomorphism if (and only if) every regular element of $R$ is a unit. Such are rings are often called "quorings."

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