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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

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    $\begingroup$ Does Isomorphism preserve countability?? $\endgroup$
    – user87543
    Dec 28 '13 at 15:40
  • $\begingroup$ @PraphullaKoushik It must preserve cardinality, as it must be a bijection. $\endgroup$
    – Brian
    Dec 28 '13 at 15:40
  • $\begingroup$ This certainly works, and I know of no better way. $\endgroup$ Dec 28 '13 at 15:42
  • $\begingroup$ @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. $\endgroup$
    – Brian
    Dec 28 '13 at 15:42
  • $\begingroup$ @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. $\endgroup$
    – John Smith
    Dec 28 '13 at 15:43
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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.

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  • $\begingroup$ This answer is very nice. I used your idea to prove this (math.stackexchange.com/questions/620551/…). $\endgroup$
    – John Smith
    Dec 28 '13 at 16:22
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    $\begingroup$ 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. $\endgroup$
    – 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.

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  • $\begingroup$ Why must the map be a vector space isomorphism? It need not fix $ \Bbb Q$. $\endgroup$ Dec 28 '13 at 15:51
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    $\begingroup$ 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 $\Bbb C$). $\endgroup$ Dec 28 '13 at 15:57
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    $\begingroup$ @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 ? $\endgroup$
    – Robert M
    Dec 28 '13 at 16:00
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    $\begingroup$ @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. $\endgroup$ Dec 28 '13 at 16:00
  • $\begingroup$ It would be helpful to explain the reason when you downvote a correct answer ! $\endgroup$
    – 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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  • $\begingroup$ The last comment above may be tough without proof for someone asking this question... $\endgroup$
    – DonAntonio
    Dec 28 '13 at 16:20
  • $\begingroup$ I have not yet covered quotient groups. It is only in chapter $3$ of Dummit and Foote. I am still in chapter $1$. $\endgroup$
    – John Smith
    Dec 28 '13 at 16:21

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