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Based on this question:

Is it true that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as abelian groups?.

I'm trying to prove that this map $\varphi:\mathbb{R}\to\mathbb{R}^2$ defined by its action on element of Hamel basis of $\mathbb R$: $\varphi(e_\lambda)= (e_{i(\lambda)_1},e_{i(\lambda)_2})$, where $i:\Lambda\to\Lambda\times\Lambda$ is a set theoretic bijection, is an isomorphism between $\mathbb R$ and $\mathbb R^2$ as $\mathbb Q$-vector spaces.

The author were focusing in another aspects of the question and leave this proof unsolved. I'm trying to understand why this is an isomorphism, I have a basic knowledge of abstract algebra, I need help.


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By the way, you should take a look at the edited verrsion of that question. Couse the first solution was wrong, though all folks said it is correct. – Norbert Jul 8 '14 at 16:55
up vote 2 down vote accepted

I'm not sure what you mean by $\Lambda$, but it appears that the map $\phi$ gives a bijection between a $\bf Q$-basis of $\bf R$ and a $\bf Q$-basis of ${\bf R}^2$, right? Now you extend $\phi$ by linearity to be a map from $\bf R$ to ${\bf R}^2$. Since you have extended it by linearity, it is a linear transformation. To show it's a $\bf Q$-vector space isomorphism, you have to show it's one-one and onto; equivalently, you have to show it's invertible.

So, given an element of ${\bf R}^2$, you can express it as a finite linear combination of the basis vectors; then can you see how to use $\phi$ to show it is the image of a unique element of $\bf R$?

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what do you mean exactly by "extend $\phi$ by linearity"? – user42912 Nov 8 '12 at 2:52
When you have a transformation $T$ defined on a basis $B$ then to extend it by linearity means this: given any $v$ in the domain, express $v$ in terms of the basis as $v=a_1b_1+\cdots+a_rb_r$ for some $b_1,\dots,b_r$ in $B$ and some scalars $a_1,\dots,a_r$, and define $T(v)=a_1T(b_1)+\cdots+a_rT(b_r)$. – Gerry Myerson Nov 8 '12 at 2:56
So suppose $v=a_1b_1+\ldots+a_nb_n$, where $\{b_1,\ldots,b_n\}$ is a basis, we define $\phi$ as$\phi(v)=a_1\phi(b_1)+\ldots+a_n\phi(b_n)$? – user42912 Nov 8 '12 at 3:06
Yes. ${}{}{}{}$ – Gerry Myerson Nov 8 '12 at 3:07
so I already proved that this function is one-one and onto, what's next? the question is done? – user42912 Nov 8 '12 at 3:17

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