The free group 𝐹𝑆 consists of all reduced words that can be built from members of 𝑆 and formal inverses of members of 𝑆.
Show the following presentation of the free abelian group of rank $n$:
$$\langle a_{1},a_{2}\cdots a_{n}\mid a_{i}a_{j}a_{i}^{-1}a_{j}^{-1} \forall1\leq i,j\leq n\rangle \cong \mathbb Z^{n}.$$
My solution:
Define $X=\{ a_{1},a_{2}\cdots a_{n}\} , S=\{ a_{i}a_{j}a_{i}^{-1}a_{j}^{-1}\}, R=\langle\langle S\rangle\rangle,G=F_X/R.$
By universal property define $$\phi : F_X \to \mathbb Z^n.$$
$$a_{i} \mapsto e_i$$
$$\phi(a_{i}a_{j}a_{i}^{-1}a_{j}^{-1})=\phi(a_{i})\phi(a_{j})\phi(a_{i}^{-1})\phi(a_{j}^{-1})=e_i+e_j-e_i-e_j=0$$
$$a_{i}a_{j}a_{i}^{-1}a_{j}^{-1}R=R \implies a_{i}^{-1}a_{j}^{-1}R=a_{j}^{-1}a_{i}^{-1}R \implies G $$ is abelian.
Since $G$ is abelian, any element can be written as $$\prod_{i=1}^{n} a_i^{k_i}=\sum_{i=1}^{n} k_ie_i.$$
$\sum_{i=1}^{n} k_ie_i = 0 \implies $each $k_i=0$ since $\{ e_i \}$ independent.
Then $ ker \phi=\{ 1\} \implies G \cong \mathbb Z^n.$
Is my proof correct?
Best Answer
After you have constructed $\phi$ using the universal property, thanks to the 1st isomrophism theorem, you only need to show two things: $\phi$ is surjective (easy: $(k_1, \cdots, k_n) = \phi(a_1^{k_1}\cdots a_n^{k_n})$, and $R=\ker\phi$. You showed that $S\subset \ker \phi$, hence $R=\langle S \rangle \subset\ker\phi$, and it's slightly harder to show $\ker\phi\subset R$. In this approach there is no reason to show $F_X/R$ is abelian.
$\ker\phi=\{1\}$ is false. You wanted to say $\tilde{\phi}:F_X/R\rightarrow \mathbb Z^n$ has trivial kernel, which is equivalent to show $R=\ker\phi$.
There is a slightly easier approach: show that $F_X/R$ is abelian and satisfies the universal property of a free abelian group with $n$ generators. You have already shown $F_X/R$ is abelian, and it's left to show for an arbitrary abelian group $A$ and map $X\rightarrow A$, there is a unique lifting $\rho: F_X/R\rightarrow A$.