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Let $V$ be a finite dimensional vector space (of dimension n) over a field $F$. I need to show that $V$ is isomorphic to $F^n$ as abelian groups. However, I don't really understand what does "isomorphic as abelian groups" mean, my (poor) attempt to solve this was:

Let $f:V\to F^n$, $f((v_1,v_2,\ldots,v_n)):=v_1+\ldots+v_n$, clearly $f$ is a group homomorphism, but it is not biyective since, for example, in the case where $dim(V)=2$ and $F=\mathbb{R}$ we have $f((2,-2))=0=f((3,-3))$, I've also tried it by defining $f$ to be the product of a vector's entries but obviously it didn't work so I'm stuck!

I think this should be an easy problem but since I don't quite understand it, it's giving me problems. Thank you for your help.

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    @ChrisEagle would you be able to take a look at this? https://math.stackexchange.com/questions/2374069/generalization-of-n-th-dimensional-vector-space-isomorphic-to-n-th-product-o I'm trying to generalize what Zero is asking about here.2017-07-27

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First, pick a basis $\{w_i\}$ for $V$. Note that all elements of $V$ can be written as $\sum^n c_iw_i$ for some $c_i \in F$. This gives a natural function $f(\sum^n c_iw_i) = (c_1, ... c_n)$. Show that it is a bijection and that it is a homomorphism as far as $+$ is concerned.

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    @KarolisJuodele I don't know if you would be able to take a look, but I'm trying to generalize this to the case where instead of $n$ copies we are looking at the direct sum of copies of $F$ indexed by the basis of $V$. https://math.stackexchange.com/questions/2374069/generalization-of-n-th-dimensional-vector-space-isomorphic-to-n-th-product-o2017-07-27