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I have to prove that the two groups $ ( \mathbb{Q}, +)$ and $ ({\mathbb{R},+ }) $ are not isomorphic.
Both of these groups are infinte ,abelian and non-cyclic, so I cannot use the fact that they are "not" isomorphic 'cause one of them is cyclic/abelian and the other is not.

Next, both of them have only one i.e., the identity element of finite order and no other is of finite order so I cannot use the contradiction method that one of them has one element of finite order and the other has none, like I used to prove the same fact for groups, $ ( \mathbb{Q}, +)$ and $ ({\mathbb{R^{*}},+ }) $.

I think I can try to use contradiction method to prove this statement but haven't got any far with that.
So , my question is: $1.$ Is there any way other than contradiction to prove this statement or statements of this type?
$2.$ How can I use contradiction method to solve this problem?

I will appreciate any kind of help or hint I can get.
Thanks in advance.

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    $\begingroup$ They have different cardinality. $\endgroup$ – Crostul Mar 20 '15 at 10:43
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Suppose $\varphi:\mathbb Q\rightarrow\mathbb R$ is a group isomorphism. Then it must be bijective, in particular it has to be surjective. It is well established that this is impossible since $|\mathbb Q|=|\mathbb N|<|\mathbb R|$.

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Hint: For a proof by contradiction note that a group isomorphism is also a set isomorphism.

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  • $\begingroup$ can you explain a little bit more? $\endgroup$ – Mojojojo Mar 20 '15 at 10:47
  • $\begingroup$ A group isomorphism $(\mathbb{Q}, +, 1) \xrightarrow[\sim]{} (\mathbb{R}, +, 1)$ is an isomorphism $\mathbb{Q} → \mathbb{R}$ with an additional property. Does an isomorphism between $\mathbb{Q}$ and $\mathbb{R}$ exist? $\endgroup$ – user3493525 Mar 20 '15 at 10:54
  • $\begingroup$ no, there does not and thanks. $\endgroup$ – Mojojojo Mar 20 '15 at 10:57

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