# Proof of the universal property of free abelian groups

Let $$S$$ be a set. The group with presentation $$(S, R)$$ , where $$R = \{\ [s , t] \mid s,t\in S\ \}$$ is called the free abelian group on $$S$$ -- denote it by $$A (S)$$ . Prove that $$A (S)$$ has the following universal property: if $$G$$ is any abelian group and $$\varphi:S\to G$$ is any set map, then there is a unique group homomorphism $$\phi : A (S) \to G$$ such that $$\phi\mid_S=\varphi$$ . Deduce that if $$A$$ is a free abelian group on a set of cardinality $$n$$ then $$> A\cong\mathbb{Z}\times\mathbb{Z}\times\cdots\times\mathbb{Z}\ \ (n\text{ factors})\ . >$$

For all $$N\unlhd G$$ containing $$R$$ , we have $$[s,t]\in R\subseteq N$$ , then $$[s,t]N=N$$ , and so \begin{aligned} \,[s,t]\langle R\rangle&=[s,t]\bigcap_{R\subseteq N\unlhd F(S)}N \\&=\bigcap_{R\subseteq N\unlhd F(S)}N \\&=\langle R\rangle\ , \end{aligned} hence \begin{aligned} st\langle R\rangle&=ts\langle R\rangle \\s\langle R\rangle\,t\langle R\rangle&=t\langle R\rangle\,s\langle R\rangle \end{aligned} implies $$F(S)/\langle R\rangle$$ is abelian.

Denote $$F(S)'$$ the commutator subgroup of $$F(S)$$ . Let $$\pi':F(S)\to F(S)/F(S)'$$ and $$\pi_R:F(S)\to F(S)/\langle R\rangle$$ be the natural projection with $$\ker\pi_R=\langle R\rangle$$ . Then by the universal property of commutator subgroups, we have the following commutative diagram:

Then by the universal property of free groups, the diagram below also commutes:

Combining them I get

But I do not know how to continue this work to obtain a group homomorphism as desired between the two abelian groups $$F(S)/\langle R\rangle$$ and $$G$$ .

• Please avoid using images to convey critical information not otherwise present in your post. Here is why. (In particular, your images show way too large in my current display). I know that MathJax isn't great for commutative diagrams (plain $\LaTeX$ isn't great either), but there are ways to do commutative diagrams here. But there are ways to do it. Commented Mar 14 at 16:24
• @ArturoMagidin Thanks for your advice, I had spent half an hour finding the way to draw diagonal arrows by AMScd in Mathjax before I posted my question, but it seems that there exists no way to draw a long diagonal arrow, so I had to use tikzcd to draw one instead. Commented Mar 15 at 0:29
• The result looks awful on at least two of my devices. Way too large, and on very noticeably different enlargement scales for each of the three images. Which is just one of the problems with images: they cannot be scaled like the rest of the website. I don't know if they look ok to you, but they definitely do not look ok to me. They make your post harder to read and understand. I suspect a short diagonal arrow would have been a much better solution. Commented Mar 15 at 1:11
• @ArturoMagidin It looks ok on my phone but it is too large on my computer. Next time I will directly show the universal property by text. Commented Mar 15 at 3:01

Lemma. The normal subgroup of $$F(S)$$ generated by $$R$$, $$\langle R\rangle^{F(S)}$$, is precisely $$[F(S),F(S)]$$.

Proof. Since $$R$$ is generated by elements of $$[F(S),F(S)]$$, then we have $$\langle R\rangle^{F(S)}\leq [F(S),F(S)]$$.

Conversely, using the commutator identities (I use the convention $$[a,b]=a^{-1}b^{-1}ab$$; similar identities hold for the other convention), we have: \begin{align*} [x,zy] &= [x,y][x,z]^y\\ [xz,y] &= [x,y]^z[z,y]\\ [x,y^{-1}] &= [y,x]^{y^{-1}}\\ [x^{-1},y] &= [y,x]^{x^{-1}} \end{align*} it follows that any element of $$[F(S),F(S)]$$ can be written as a product of conjugates of elements of the form $$[s,t]$$ with $$s,t\in S$$, so $$[F(S),F(S)]\leq \langle R\rangle^{F(S)}$$, giving equality.

(Alternatively, you already proved that $$[F(S),F(S)]\leq \langle R\rangle^{F(S)}$$ using the universal property of the abelianization, but I figured one universal property at a time is enough). $$\Box$$

Proof of the Universal Property.

Existence. Let $$G$$ be an abelian group, and let $$\varphi\colon S\to G$$ be a set map. Then there exists a (unique) morphism $$\Phi\colon F(S)\to G$$ such that $$\Phi\circ i = \varphi$$, where $$i\colon S\to F(S)$$ is the inclusion of $$S$$ to the free group on $$S$$.

Now we have a morphism $$\Phi\colon F(S)\to G$$, where $$G$$ is abelian. Therefore, there exists $$\Psi\colon F(S)/[F(S),F(S)]\to G$$ such that $$\Phi=\Psi\circ\pi$$, where $$\pi\colon F(S) \to F(S)/[F(S),F(S)]$$ is the canonical projection. But $$[F(S),F(S)]=\langle R\rangle^{F(S)}$$, so the domain of $$\Psi$$ is actually $$A(S)$$. Note that the inclusion of $$S$$ into $$A(S)$$ is $$\pi\circ i$$. Thus, we have a morphism $$\Psi\colon A(S)\to G$$ such that $$\Psi|_S = \Psi\circ (\pi\circ i) = (\Psi\circ \pi)\circ i = \Phi\circ i = \varphi,$$ as desired.

Uniqueness. If $$F\colon A(S)\to G$$ is such that $$F\circ (\pi\circ i) = \varphi$$, then $$F$$ and $$\Psi$$ agree on $$S$$; since $$A(S)$$ is generated by $$S$$, it follows that $$F=\Psi$$, as required. $$\Box$$

• Oh, you are knowledgeable in group theory. But I don't understand the superscript $y$ in $[x, yz]=[x, y][x, z]^y$ in the proof of your lemma. Commented Mar 15 at 1:00
• @DianWei $x^g=g^{-1}xg$; ${}^{g}x=gxg^{-1}$. Commented Mar 15 at 1:04
• There is a typo in $[x,yz]=[x,y][x,z]^y$ , replace $[x,yz]$ by $[x,zy]$ then it truly holds. Commented Mar 15 at 2:57
• Your statement reminds me that $\langle R\rangle$ is just the closure by subgroups, and $\langle R\rangle^{F(S)}$ is the closure by normal subgroups, they are not the same in the case that $\#S=2$ : $\langle R\rangle$ is finitely generated but $\langle R\rangle^{F(S)}$ is not. Commented Mar 15 at 3:08
• @DianWei Typo fixed. As to the other comment, I do not see how it is relevant to my post or why you decided to add it here. Commented Mar 15 at 3:38