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My professor didn’t define what is a free abelian group $G$ (on a set $X$) but I can guess out the definition.

We can define a free abelian group $G$ on a set $X$ as a free object on $X$ in the category of abelian groups. I don’t know if my definition coincides with the mainstream definition though.

From this post, we know, strictly speaking, abelian groups and $\mathbb{Z}$-modules are not the same because Abelian groups are sets with one operation and $\mathbb{Z}$-modules are sets with two operations. However, they are isomorphic in the sense that the categories of abelian groups and $\mathbb{Z}$-modules are isomorphic.

Then two questions about the definition of free abelian groups naturally arose in my mind.

Q1: I saw people were talking about free abelian groups across this site, e.g., here, without an underlying set. I can only define a free abelian group on some set $X$. How do we define a free abelian group without an underlying set?

Q2: I also saw people were saying that free abelian groups are the same with free $\mathbb{Z}$-modules. From the point of view of set theory, they are not the same as we mentioned above. Do people actually mean that free abelian groups are isomorphic to free $\mathbb{Z}$-modules?

Thanks for help. Sorry for my long description but I just wanted to state things clear.

Jendrik Stelzner
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Sam Wong
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  • A free abelian group of rank $n$ is $\Bbb Z^n$. Of these only $\Bbb Z$ is free. – suckling pig Dec 26 '22 at 05:07
  • Probably if you let $n=0$ you get the trivial group. It's free too. Not sure about the module stuff. – suckling pig Dec 26 '22 at 05:28
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    @eggnog Yea, your definition is a concrete one (which is equivalent to my definition, cf. this post, some knowledge of category theory is needed for reading the post). The rank in your definition is actually the cardinality of the underlying set $X$ in my definition. Anyway, go bears! – Sam Wong Dec 26 '22 at 05:34
  • Yes, all that seems to matter is $X$'s cardinality. – suckling pig Dec 26 '22 at 05:54

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Q1 For every set $X$, there is an abelian group called "the free abelian group on $X$", which can be constructed explicitly as the group of formal $\mathbb{Z}$-linear combinations of elements of $X$.

If $A$ is an abelian group, we say that $A$ is "free" if there exists a set $X$ such that $A$ is isomorphic to the free abelian group on $X$.

Some abelian groups are free (e.g. the trivial group, $\mathbb{Z}$, $\mathbb{Z}^2$), and some are not (e.g. $\mathbb{Z}/2\mathbb{Z}$).

Q2 What this means is that the isomorphism $F$ between the category of abelian groups and the category of $\mathbb{Z}$-modules has the property that $A$ is free if and only if $F(A)$ is free. In other words, this isomorphism restricts to an isomorphism between the categories of free $\mathbb{Z}$-modules and free abelian groups. Or in other words, when we pretend that "abelian group" and "$\mathbb{Z}$-module" are the same thing, then the two definitions of "free" also agree.

diracdeltafunk
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  • $\Bbb Z^2$ isn't free. It's free abelian though. – suckling pig Dec 26 '22 at 05:50
  • $\mathbb{Z}^2$ is not a free group, but it is a free abelian group. The adjective has distinct meanings in these two contexts. It is confusing! – diracdeltafunk Dec 26 '22 at 05:51
  • @diracdeltafunk We can regard $\mathbb Z^2$ as a free group modulo the commutativity, right? i.e. We identify $xy$ with $yx$ in the underlying generating set ${x,y}$ for $\mathbb Z^2$. – Sam Wong Dec 26 '22 at 06:04
  • Because of the commutativity $\Bbb Z^2$ isn't free. The free group on two generators is $F_2$. Otoh $F_1=\Bbb Z$. – suckling pig Dec 26 '22 at 06:07
  • @eggnog Yea, $\mathbb Z^2$ is not free. What I wanted to say was that, $\mathbb Z^2 \cong F_2/\sim$, where $\sim$ is an equivalent relationship: $x\sim y$ if and only if $xy=yx$, where $x,y$ are the generators of $F_2$. The notation $F_2/\sim$ means the set consisting of equivalent classes under the equivalent relationship $\sim$ in $F_2$. Then $F_2/\sim$ is a group with the addition operation induced by the one in $F_2$. – Sam Wong Dec 26 '22 at 06:20
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    This is not exactly a coherent description of the equivalence relation, but you have the right idea. If you take a free group on $n$ generators, and "abelianize it" by quotienting out by the equivalence relation you have in mind, you produce the free abelian group of rank $n$. – diracdeltafunk Dec 26 '22 at 06:31