Free groups and Their Subgroups
Theorem
Let $F$ be a free Abelian group.
- [1]: Every subgroup $R$ of $F$ is a free group.
- [2]: If $F$ has rank $n$, then the subgroup $R \subset F$ of $F$ is a free Abelian group with rank $r \le n$.
- [3]: Moreover, there exists a basis $e_{1} , \cdots , e_{n}\in F$ and natural numbers $t_{1} , \cdots , t_{k}$ for $F$ that satisfy the following three conditions:
- (i): $k \le r$, and for all $i$, $t_{i} > 1$.
- (ii): $t_{1}e_{1} , \cdots , t_{k}e_{k} , e_{k+1} , \cdots , e_{r}$ is a basis of $R$.
- (iii): $t_{1} \mid t_{2} \mid \cdots \mid t_{k}$. In other words, $t_{i}$ divides $t_{i+1}$.
While there’s no guarantee that $e_{1}, \cdots, e_{n}$ is unique, the $t_{1} , \cdots , t_{k} > 1$s are uniquely determined by the given $F$ and $R$.
Explanation
The $t_{1} , \cdots , t_{k}$ mentioned in the theorem might seem complex, but in fact, without needing to add explanations like $k \le r$ or $t_{i} > 1$, $$ t_{1}e_{1} , \cdots , t_{r}e_{r} $$ is the basis of $R$, and if we say $t_{1} , \cdots , t_{r} \in \mathbb{N}$, it’s actually clearer. However, we start this way because, like the fundamental theorem of finitely generated Abelian groups related to algebraic topology, we end up distinguishing $k \le r$.
Proof
[1], [2] 1
The projection map is defined to show the details explicitly. For a detailed proof, refer to Munkres1.
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[3] 2
According to theorems [1], [2], the subgroup $R \subset F$ of the free Abelian group $F$ with rank $n$ is a free Abelian group, and its rank is $r \le n$. Since $R$ is a subgroup of $F$, define an operation-preserving homomorphism $f : R \to F$, and choose the bases $a_{1} , \cdots , a_{r}$ for $R$ and $e_{1} , \cdots , e_{n}$ for $F$.
Homomorphism with Smith Normal Form: If the free Abelian groups $G$, $G'$ have ranks $n,m$ and $f : G \to G'$ respectively, and if $g$ is a homomorphism, then there exists a homomorphism $g$ with the following matrix: $$ \begin{bmatrix} d_{1} & 0 & 0 & 0 & \cdots & 0 \\ 0 & \ddots & 0 & 0 & \cdots & 0 \\ 0 & 0 & d_{r} & 0 & \cdots & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 0 \end{bmatrix} \in \mathbb{Z}^{m \times n} $$ where $d_{1} , \cdots, d_{r} \in \mathbb{N}$ and $d_{1} \mid \cdots \mid d_{r}$, meaning $d_{k}$ must be a divisor of $d_{k+1}$.
Since $f$ is an injection, i.e., monomorphism, the matrix $\left( \lambda_{ij} \right)$ of $f$ cannot have a zero column. For this homomorphism $f$, there exists a homomorphism $g : R \to F$ with the matrix $$ \begin{bmatrix} d_{1} & 0 & 0 & 0 & \cdots & 0 \\ 0 & \ddots & 0 & 0 & \cdots & 0 \\ 0 & 0 & d_{r} & 0 & \cdots & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 0 \end{bmatrix} $$ specifically for $i = 1 , \cdots , r$, $$ g \left( a_{i} \right) = d_{i} e_{i} $$ Therefore, since $g \left( a_{i} \right) = a_{i}$, $b_{1} e_{1} , \cdots , b_{r} e_{r}$ is the basis of $R$.
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