Prove or disprove: $\mathbb{Q}$ is isomorphic to $\mathbb{Z} \times \mathbb{Z}$. I mean the groups $(\mathbb Q, +)$ and $(\mathbb Z \times \mathbb Z,+).$ Is there an isomorphism?
Prove or disprove: $(\mathbb{Q}, +)$ is isomorphic to $(\mathbb{Z} \times \mathbb{Z}, +)$?
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8Isomorphic as what? Fields? Rings? $\mathbb{Z}$-modules? – 2012-05-17
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0@Neal One of the tags is "group-theory", so "isomorphic as groups" is probably what's implied – 2012-05-17
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7Welcome to math.SE: since you are a new user, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Also, many find the use of imperative ("Prove") to be rude when asking for help; please consider rewriting your post. – 2012-05-17
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0I mean the groups (Q,+) and (ZxZ,+). Is there an isomorphism ? – 2012-05-17
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12Let $\phi: \mathbb{Q}\to \mathbb{Z}\times \mathbb{Z}$ be an isomorphism, and suppose that $\phi(a)=(1,0)$. Then $\phi(\frac{a}{2}+\frac{a}{2})=\phi(\frac{a}{2})+\phi(\frac{a}{2})=(1,0)$. However, there is no element $u$ of $\mathbb{Z}\times \mathbb{Z}$ such that $u+u=(1,0)$. – 2012-05-17
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0@AndréNicolas What is $\phi(a)$ : $\forall $ a in Q ? – 2012-05-17
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1@AHH: There is no such isomorphism $\phi$, that is what is being proved. **If** there was an isomorphism $\phi$, then *some* element of $\mathbb{Q}$ would be mapped by $\phi$ to $(1,0)$. My comment shows that there cannot be such an $a$, since if there was such an $a$, $\phi(a/2)$ would have to be some $u$ such that $u+u=(1,0)$, and there is no such $u$. A proof along the same lines has now been posed by DonAntonio. – 2012-05-17
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3AHH My answer and @André's is basically saying that there isn't an isomorphism. *If* such an isomorphism were to exist, *then* that leads to a contradiction. You don't need to know exactly what $\phi$ is for every element in $\mathbb{Q}$, you just need *one* thing that doesn't make sense. – 2012-05-17
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0AndréNicolas , @Thomas: Are you mean the following: Let $ m/n \in \mathbb{Q} , (m,n) \in \mathbb{Z} $ ,then If we have $ m/n \mapsto (m,n) $, then $ m/1 \mapsto (m,1) $ and $m/2 \mapsto (m,2)$ $ \Rightarrow m/2 + m/2 \mapsto (2m,4) $ but $2m \neq m $ and $ 4 \neq 1 $ "Contradiction". – 2012-05-17
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0@Neal I think the answers here work for the rational numbers under normal addition (I only understand Andre Nicolas's answer), and all the pairs of integers under normal addition. We do *not* need to know anything more about such structures. As both additions qualify as the only binary operations here, we only need to know whether or not both sets have the same cardinality (equivalently a bijection exists), and whether or not there exists a function F which satisfies the relevant homomorphic equality F(a+b)=(F(a)+'F(b)), where "+" indicates one of the additions and "+'" the other addition. – 2012-05-17
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0@DougSpoonwood Is | $\mathbb{R}^2 $ | = |$\mathbb{C}$| ? – 2012-05-17
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0@Neal continued... It shouldn't come as too hard to set up a finite operation table F' say for structure (*F*, F) where *F* indicates the set, and F the binary operation, which does NOT satisfy commutavity, associativity, does not have a zero, is not idempotent, does not have an identity, etc. and thus has no label in the lexicon of abstract algebra other than "magma" (although you could make up axioms for it if you like). Then you can set up an isomorphism to find the other table, or find the other table to find an isomorphism... either way. – 2012-05-17
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0@AHH I didn't know and did a search. It looks like the answer is "yes" (since the reals R, and RxR have the same cardinality) from this source: http://answers.yahoo.com/question/index?qid=20090420075438AABM71T, but I don't know any details. – 2012-05-17
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0@DougSpoonwood, The cardinality is important when it is finite.For example we know $ \mathbb{R}^+ \varsubsetneqq \mathbb{R}$ and $ \mathbb{R}^+ \cong \mathbb{R} $. – 2012-05-17
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0@AHH I have no idea why you're saying things like that or how it bears relevance to this question. The cardinality is important both in the infinite and finite cases. – 2012-05-17
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0@Neal I should add that the function F which satisfies the relevant homomorphic equality is also the bijection. – 2012-05-17
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2@DougSpoonwood: In general, if $X$ is an infinite set and $n$ is some finite number then $|X^n|=|X|$ (it is sufficient to prove $|X^2|=X$). It is only when you try to do things like $X^{\aleph_0}$ that you go "up" a level or too. However, the fact that $\mathbb{R}^2=\{(a, b): a, b\in\mathbb{R}\}$ and $\mathbb{C}=\{a+ib: a, b\in\mathbb{R}\}$ have the same cardinality is surely easy, as the map $(a, b)\mapsto a+ib$ is a bijection... – 2012-05-17
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0@user1729 Thanks! – 2012-05-17
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0Okay, thanks @DanielFreedman and @DougSpoonwood! – 2012-05-18
7 Answers
Yet another way to see the two cannot be isomorphic as additive groups: if $a,b\in\mathbb{Q}$, and neither $a$ nor $b$ are equal to $0$, then $\langle a\rangle\cap\langle b\rangle\neq\{0\}$; that is, any two nontrivial subgroups intersect nontrivially. To see this, write $a=\frac{r}{s}$, $b=\frac{u}{v}$, with $r,s,u,v\in\mathbb{Z}$, $\gcd(r,s)=\gcd(u,v)=1$. Then $(su)a = (rv)b\neq 0$ lies in the intersection, so the intersection is nontrivial.
However, in $\mathbb{Z}\times\mathbb{Z}$, the elements $(1,0)$ and $(0,1)$ are both nontrivial, but $\langle (1,0)\rangle\cap\langle (0,1)\rangle = \{(0,0)\}$.
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0$\langle a\rangle\cap\langle b\rangle\neq\{0\}$ $ \implies \{0\} \langle a\rangle$ and $\{0\} \langle b\rangle $ $ \implies \langle a\rangle$ and $\langle b\rangle$ are not subgroups of $\mathbb{Q}$. – 2012-05-17
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0@AHH: No; $\langle a\rangle\cap\langle b\rangle\neq\{0\}$ means that the intersection **is not** just $0$. That could be *either* because $0$ is not in the intersection, or, in this case, because therre are things **other than $0$** that are *also* in the intersection. It does not mean that $0$ is not in the intersection. **By definition**, $\langle a\rangle$ means "the smallest subgroup that contains $a$", so it **must** be a subgroup. – 2012-05-17
I assume that you are asking whether we have an isomorphism of additive groups.
In that case, assume that $\phi: \mathbb{Q} \to \mathbb{Z}\times \mathbb{Z}$ is such an isomorphism. So we have for example that $\phi(0) = (0,0)$. Let $a\in \mathbb{Q}$ be such that $\phi(a) = (1,0)$ and $b$ be such that $\phi(b) = (0,1)$. Then we see that $\mathbb{Q}$ is equal to $\{na + mb \lvert n,m \in \mathbb{Z}\}$. This is a contradiction...
(Hence the argument is that $\mathbb{Q}$ is not finitely generated while $\mathbb{Z}\times \mathbb{Z}$ is.)
(I will leave the details to you.)
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0Note that: $\forall a \in \mathbf{Q} \implies a = m/n : m,n \in \mathbf{Z} , n \neq zero.$ – 2012-05-17
Another argument that you can construct with the following (be sure you can prove/answer every section):
1) An abelian additive group $\,A\,$ is said to be divisible if $\,\forall a\in A\,\,n\in\mathbb{N}\,\,\exists b\in A\,\,s.t.\,\,nb=a\,$ . To be sure, $n\neq 0$
2) $\,\mathbb{Q}\,$ is a divisible group
3) Any homomorphic image of a divisible group is a divisible group
4) Is $\,\mathbb{Z}\times\mathbb{Z}\,$ divisible?
Let $ \phi: \mathbb{Q} \to \mathbb{Z}\times \mathbb{Z}$ be a homomorphism. Fix $u/v\in\mathbb{Q}$ and let $(a_n,b_n)=\phi(u/v^n)$. Since $\phi(u/v)=v^{n-1}\phi(u/v^n)$, we get $a_1=v^{n-1}a_n$ and $b_1=v^{n-1}b_n$ for all $n\in\mathbb N$, which is clearly impossible unless $\phi(u/v)=(a_1,b_1)=(0,0)$.
So, the only homomorphism $\mathbb{Q} \to \mathbb{Z}\times \mathbb{Z}$ is the zero map, and there is no chance of an isomorphism.
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0Can we find $\phi: \mathbb{Q} \to \mathbb{Z}\times {0} \varsubsetneqq \mathbb{Z}\times \mathbb{Z} $ "homomrphism"?! – 2012-05-17
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0@AHH, no, by the same reason. – 2012-05-17
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0if $\phi(a)= \lfloor a \rfloor ,\forall a \in \mathbb{Q} $, then $\phi$ a homomrphism. – 2012-05-17
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0@AHH, no, it's not: $\phi(1/2)+\phi(1/2)=0$ but $\phi(1)=1$. – 2012-05-17
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0good counterexample :) – 2012-05-17
Another idea: suppose there is an isomorphism $\mathbb Q \to \mathbb Z \oplus \mathbb Z$, then tensor both sides $\otimes_\mathbb{Z} \mathbb Q$, and get a $\mathbb Q$-module isomorphism $\mathbb Q = \mathbb Q \otimes_\mathbb{Z} \mathbb Q \to (\mathbb Z \oplus \mathbb Z)\otimes_\mathbb{Z} \mathbb Q = \mathbb Q\oplus \mathbb Q$.
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0This also works with $\otimes_\mathbb{Z} \mathbb Z /n$ for any $n\ge 2$, giving an isomorphism $0 \to \mathbb Z/n \oplus \mathbb Z/n$, which of course is just a rephrasing of some of the arguments elsewhere. I like applying functors and seeing things are different that way, it's the algebraic topologist in me. – 2012-05-19
Another way of seeing this (yes, there are many ways!) is to notice that two isomorphic groups have the same quotients. That is, if $G\cong H$ and $G\twoheadrightarrow K$ then $H \twoheadrightarrow K$.
Now, by this question (which was only asked the other day, which is why I am posting this answer!), we know that every proper quotient of $\mathbb{Q}$ is torsion (that is, every element has finite order). On the other hand, $\mathbb{Z}\times\mathbb{Z}$ has a torsion-free proper quotient, $\mathbb{Z}\times\mathbb{Z}\twoheadrightarrow \mathbb{Z}$. Thus, they cannot be isomorphic.
(Indeed, this actually proves that there cannot be a homomorphism from $\mathbb{Q}$ to $\mathbb{Z}\times\mathbb{Z}$, as lhf has already shown - the result we use tells us that the map cannot have non-trivial kernel, while this proves that the kernel cannot be trivial either.)
You can show that the group $(\mathbb{Q}, +)$ has no proper subgroup of finite index, but for example $\mathbb{Z} \times 2\mathbb{Z}$ has finite index in ($\mathbb{Z} \times \mathbb{Z}, +)$.
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0That's not true - every proper, non-trivial subgroup of $(\mathbb{Q}, +)$ has finite index! (Remember, every quotient is finite.) – 2012-05-17
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0@user1729: If $M$ is a proper subgroup of finite index $d$, then for any $q \in \mathbb{Q}$ you have $d(q + M) = M$ so $dq \in M$. But take some $r \not\in M$ and you get $d(r/d) \not\in M$, a contradiction. – 2012-05-17
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0Sorry - my brain is slow this morning... – 2012-05-17