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I am kind of lost when I try to understand the definition of the free abelian group on a set $X$

I read the question here What is the definition of a free abelian group

According to this question, the free abelian group of $\{A,B,C,D\}$ is $\mathbb{Z}^4$. However, I think that they have to be linearly independent to make that group isomorphic to $\mathbb{Z}^4$. For example, what if there is an implicit relation $2A+B=0$. Then obviously it is not $\mathbb{Z}^4$, since $(2,1,0,0)$ and $(4,2,0,0)$ are the same thing. What is wrong with my argument?

I also read Serge Lang's algebra book. It seems that if we say a free abelian group on $X$, then we assume $X$ form a basis. All elements of $X$ are linearly independent.

I am confused. For example, if we take the free abelian group on $\{1,2,3\}$, is it still $\mathbb{Z}^{3}$ (instead of $\mathbb{Z}$)?

$2\times1+(-1)\times2+1\times3=4\times1+(-2)\times2+1\times3$, but still $(2,-1,1)$ and $(4,-2,1)$ are diffrent?

MathEric
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    If there's an implicit relationship between two of the generators, then you're no longer talking about the free abelian group on those generators. – Robert Shore Aug 27 '19 at 00:28

2 Answers2

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One way to think about the free abelian group on a set $X$ is that you are taking the elements of $X$ as a starting point, and then making a group operation on those elements that is abelian, but where the elements don't satisfy any relations other than the ones that they absolutely have to to make an abelian group. In particular, the elements of $X$ are viewed here as something like "atoms", and we're building a totally new group operation on them. It doesn't matter at all what they are, just how many of them there are.

In your second example the fact that we built a group from the numbers $1, 2, 3$ doesn't mean that the group operation we're building has anything to do with the usual arithmetic on those numbers. It might help to give the new group operation a new name, say $\ast$. Then elements of the free abelian group on $\{1, 2, 3\}$ will have the form $(a1)\ast(b2)\ast(c3)$, with $a, b, c \in \mathbb{Z}$, and with no further simplification possible. The group we get this way is isomorphic to $\mathbb{Z}^3$, via the function that sends $(a1)\ast(b2)\ast(c3)$ to $(a, b, c)$.

Chris Eagle
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A more formal way: let $F(S)$ be the free abelian group on a set $S$. (In your case $S=\{A,B,C,D\}$.) If $S'$ is another set which is in bijection to $S$, then $F(S)$ and $F(S')$ are isomorphic, the isomorphism being induced by the bijection.

Loco
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