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Show that $(\Bbb Q, +)$ and $(\Bbb R, +)$ are not isomorphic groups.

I dont how to proceed here. My general strategies include: trying to show that if one group has an element of a particular order then the other group does not have a corresponding element of that particular order, showing that we can't find a pre-image of the "image" group. But I don’t know what to do here? Also, I am still learning elementary group theory and I need an "elementary" solution for the same (without using cardinality of sets, as I still don't know about countability, uncountablity and other related topics).

A good answer is provided by Alex J Best. But I still need a little clarification. A property true for rationals is used to solve the problem mentioned. The property goes as follows:

For every $x,y\in \Bbb Q$ nonzero there exists integers $m, n \ne 0$ such that $mx = ny$.

I proved this property as follows:

The proof of this property for rationals, follows from the definition of a rational number since, if $k=\frac yx\in\Bbb Q$ (,where $\exists x,y\in\Bbb Z$ such that $gcd(x, y) = 1$ and $x\neq 0$), then we have, $m,n\neq 0$, (satisying $m, n\in\Bbb Z$), such that $\frac mn=\frac yx$(if, $m=ky,n=kx$, then it is true evidently with $k\neq 0$) and the result follows immediately.

Is this correct? If not, where is it going wrong?...

This link Prove that the additive groups $\mathbb{R}$ and $\mathbb{Q}$ are not isomorphic. does not answer my question because: All the tools utilised to answer this question in the thread, are not at all elementary (at least to me). Actually, in our course, we have not been taught about concepts like vector spaces, Cantor's diagonal including the ones written previously in this post. So that is not answering my question.

Arthur
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  • Shouldn't it suffice to show that $\mathbb{Q}$ and $\mathbb{R}$ are respectively incomplete and complete metric spaces, and since the operation is the same (+) they cannot yield isomorphic groups? – Superunknown Jan 05 '23 at 12:32
  • @DietrichBurde Thank you! I am so much grateful. I haven't noticed this duplicate at all. But still , this is not answering my question because all the tools utilised to answer this question in the thread, are not at all elementary (atleast to me). Actually, in our course, we have not been taught about concepts like vector spaces, Cantor's diagonal including the ones written in the post. So that is not answering my question . – Arthur Jan 05 '23 at 13:05
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    @DietrichBurde I understand you , and that maybe true of course . But still, that's the way it's been taught here. I am following Herstein's book for group theory and those concepts are not utilised up until the chapter, where I am in now. You are probably, more than just correct , but still, I am not seeing a better way than posting this in my case, as it is of now. But thank you so much, for your valuable suggestions. – Arthur Jan 05 '23 at 13:14
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    I think you should give the "best" solution a chance here. Certainly cardinality is answering your question immediately and in the most natural way. Instead of avoiding this now, as you do, you should rather start learning about cardinality (takes only a few minutes and its very basic, and you will need it anyway for everything in math). And I bet, the exercise is just for learning about cardinality (as a self study). – Dietrich Burde Jan 05 '23 at 13:17
  • @DietrichBurde I don't know, if I can make this request to you,( or rather , if I am at all eligible to make this request), but can you suggest me a source(or maybe a handout or a reference) for the same , that would be very much helpful. – Arthur Jan 05 '23 at 13:19
  • @EthanBolker Thank you so much for your suggestion but if you please go through the comments , I provided an explanation why it does not answer my question . Should I copy-paste those reasons in my post , what do you suggest? – Arthur Jan 05 '23 at 13:22
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    If you are reading Herstein's book, then everything suggested in these comments regarding cardinality that is needed to answer your question is right there, in Chapter 1. You should pay particular attention to the sections entitled "Sets" and "Mappings". Don't skip Chapter 1. – Lee Mosher Jan 05 '23 at 14:45

2 Answers2

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One elementary property of groups that is satisfied by $\mathbf Q$ but not by $\mathbf R$ is the following:

For every $x,y\in G$ nonzero there exists integers $m, n \ne 0$ such that $mx = ny$.

By proving that the rationals have this property but the reals do not we can see that the groups are not isomorphic.

Alex J Best
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  • @Alex J Best That's a good one!... – Arthur Jan 05 '23 at 13:07
  • @Alex J Best But can you please provide a little more elaboration ?.. – Arthur Jan 05 '23 at 13:21
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    @Franklin, which part is unclear? I purposefully didn't spell out every detail, so there is still some work to do to make this a complete solution for sure! – Alex J Best Jan 05 '23 at 14:22
  • Actually, $my=nx$, is a property true for rationals, which follows from the definition of a rational number as, if $k=\frac xy\in \Bbb Q$(, where $\exists x,y\in \Bbb Z$ such that $gcd(x,y)=1$ and $x\neq 0$), then, we have, $m,n\neq 0$, (satisying $m,n\in\Bbb Z$), such that $\frac mn=\frac xy$(if, $m=kx,n=ky$, then it is true evidently with $k\neq 0$) and the result follows immediately. Is this correct? If not, where is it going wrong?... – Arthur Jan 06 '23 at 12:28
  • Is it correct? (I mean the proof given in my previous comment) – Arthur Feb 11 '23 at 11:51
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Lemma. For any two non-zero rationals $x,y$, there exists non-zero integers $m$ and $n$ such that $mx=ny$.

Proof: $x=p/q$, $y=r/s$. So $x=\frac{ps}{qr}y$. Hence, $(qr)x=(ps)y$.

If $f:\mathbb{Q}\rightarrow \mathbb{R}$ is a isomorphism then $\mathbb{R}$ satisfies the lemma.

To see this, let $x',y'$ be non-zero reals. Choose $x$,$y$ rationals such that $f(x)=x'$ and $f(y)=y'$. Choose integers $m,n$ such that $mx=ny$. Then $mf(x)=nf(y)$. So, $mx'=ny'$.

Since $e$ is not a rational number this is a contradiction.

Some Person
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