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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?2011-11-30
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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.2011-11-30
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    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.2011-11-30
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    @Arturo Magidin : This is exactly the situation i'm in: $f$ having a retraction. i can see by using the iso theorem that $G_2\cong Im(f)\oplus ker(h)$ but why does this imply that $f$ is an iso?2011-11-30
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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.2011-11-30
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    you are absolutely right but i thought it was a property of homomorphisms on $\mathbb Z\oplus \mathbb Z$ but now i see that it is a general fact for homomorphisms with retraction.2011-11-30
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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.2011-11-30

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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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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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    Kind Sir, could you please elaborate on $ad - bc = \pm 1$ being necessary and sufficient. How could I go about proving that? I see that if it is nonzero then it is injective. Thank you2016-10-16
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    $f(x,y) = x (a,b) + y (c,d) = (x,y) \pmatrix{a & b\cr c & d}$. $f$ is an isomorphism iff the matrix $\pmatrix{a & b\cr c & d\cr}$ has an inverse with integer entries, and that is true iff the determinant is $\pm 1$.2016-10-19
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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 $2$".2011-11-30
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    @Arturo Sorry, it seems that I misunderstood your comments. Hopefully added text together with your comments clarifies the situation.)2011-11-30
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    You didn't misunderstand them: I expressed myself poorly (then deleted and rephrased). So no worries.2011-11-30
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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.