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Is there a better (or other) way(s) to prove the following statement? Also, the same argument works for multiplicative groups $\mathbb{R}-\{0\}$ and $\mathbb{Q}-\{0\}$, right?

Problem Prove that the additive groups $\mathbb{R}$ and $\mathbb{Q}$ are not isomorphic.

Solution By cantor's diagonal argument, there is no possible bijection between $\mathbb{Q}$ and $\mathbb{R}$. Since an isomorphism needs to be a bijection, there is no possible isomorphism between the additive groups $\mathbb{R}$ and $\mathbb{Q}$.

Thanks

John Smith
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    Does Isomorphism preserve countability?? –  Dec 28 '13 at 15:40
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    @PraphullaKoushik It must preserve cardinality, as it must be a bijection. – Brian Dec 28 '13 at 15:40
  • This certainly works, and I know of no better way. – PVAL-inactive Dec 28 '13 at 15:42
  • @PraphullaKoushik Oh, sorry. I suppose OP already knows this as it was listed as their reasoning in their post. I believe the question is asking not if the proof is correct, but if there is some other way to prove this. – Brian Dec 28 '13 at 15:42
  • @PraphullaKoushik I do not understand your question. Isomorphism has to be a bijection. Hence, if two groups are isomorphic, then they must have the same cardinality. I know this. I am wondering if there are other ways to prove this fact. – John Smith Dec 28 '13 at 15:43

4 Answers4

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Let $\Phi: \mathbb{Q} \rightarrow \mathbb{R}$ homomorphism of additive groups. Then $\Phi$ is already determined by $\Phi(1)$ (as $\Phi(\frac{a}{b})= \Phi(\frac{1}{b})+ ... + \Phi(\frac{1}{b})$ ($a$ summands) and $\Phi(1)= \Phi(\frac{1}{b}) + ... + \Phi(\frac{1}{b}) $, ($b$ summands))

Now, say $\sqrt{2} \cdot \Phi(1)$ doesn't have a preimage.

Edit: I just saw this was remarked by @Robert M in a comment to his answer.

Louis
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  • This answer is very nice. I used your idea to prove this (http://math.stackexchange.com/questions/620551/prove-that-the-additive-groups-mathbbz-and-mathbbq-are-not-isomorphic). – John Smith Dec 28 '13 at 16:22
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    You can simplify at the other question by using that for a homomorphism $\Phi: \mathbb{Z} \rightarrow \mathbb{Q}$, say $\frac{1}{2} \Phi(1)$ will not have a preimage. – Louis Dec 28 '13 at 16:31
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This is another approach: If $(\Bbb{R},+) \cong (\Bbb{Q},+)$ then $\Bbb{R} \cong \Bbb{Q}$ as vector spaces over $\Bbb{Q}$ which is impossible since one is one dimensional and the other is infinite dimensional.

Robert M
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  • Why must the map be a vector space isomorphism? It need not fix $ \Bbb Q$. – PVAL-inactive Dec 28 '13 at 15:51
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    In fact there are non-isomorphic vector spaces that are isomorphic not only as groups but as fields (the algebraic closure of $\Bbb C(x)$ and $\Bbb C$). – PVAL-inactive Dec 28 '13 at 15:57
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    @PVAL If $f(a + b) = f(a) + f(b)$ then $f(mz) = mf(z)$. Let $z = \frac xm$ then we drive $f(\frac xm) = \frac {1}{m}f(x)$. Now let $x = nt$ and we get $f(\frac{n}{m}t) = \frac {1}{m}f(nt) = \frac{n}{m}f(t)$. Satisfied ? – Robert M Dec 28 '13 at 16:00
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    @PVAL: any abelian group has at most one structure of a $\mathbf Q$-vector space. If two abelian groups admit the structure of a $\mathbf Q$-vector space, then homomorphisms between them are automatically $\mathbf Q$-linear. So if two torsion-free injective abelian groups are isomorphic as abelian groups, they are isomorphic as $\mathbf Q$-vector spaces. – Daniel Miller Dec 28 '13 at 16:00
  • It would be helpful to explain the reason when you downvote a correct answer ! – Robert M Dec 28 '13 at 16:03
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The quotient of $\mathbb{Q}$ by a cyclic subgroup (namely $\mathbb{Z}$) is torsion. $\mathbb{R}$ has no such subgroup.

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Let $F : Q → R$ be isomorphism of additive groups.

$n^\frac{1}{2}$ belongs to $\Bbb R$ for all $n$ natural number

Since $F$ is isomorphism, there exist unique $\frac{p}{q}$ in $Q$ for unique $n$ such that $F(\frac{p}{q}) = n^\frac{1}{2}$

$$ F\left(\frac{p}{q}\right) = F\left(\frac{1}{q} + \frac{1}{q} + \ldots + \frac{1}{q}\right) = p\cdot F\left(\frac{1}{q}\right) = n^\frac{1}{2}\qquad\qquad\mathbf{(1)} $$

$$F(1) = F\left(\frac{q}{q}\right) = q \cdot F\left(\frac{1}{q}\right) = q \cdot \frac{n^\frac{1}{2}}{p}\qquad\qquad\mathbf{(2)}$$

Hence,

$$F(1) = \frac{q_1 \cdot (1)^\frac{1}{2}}{p_1} = \frac{q_2 \cdot (2)^\frac{1}{2}}{p_2} = \frac{q_3 \cdot (3)^\frac{1}{2}}{p_3} =\ldots$$

So, $\frac{q_1 \cdot p_2}{p_1 \cdot q_2} = (2)^\frac{1}{2}$ where LHS is rational and RHS irrational

This imply our assumption is incorrect. Therefore, there doesn't exist an isomorphism $F$.

Arturo Magidin
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