This answer is devoted to the connection between *commutative groups* and *linear algebra*. It especially focuses on the link between *exponent* and *minimal polynomial* of an endomorphism, as the original poster rightfully find useful to connect notions between them.

**Definition.** Let $R$ be a commutative ring with unity, then a $R$-module is a triple $(M,+,\cdot)$, where $(M,+)$ is a commutative group and $\cdot\colon R\times M\rightarrow M$ is such that the following properties hold:
$$\begin{align}r\cdot(x+y)&=r\cdot x+r\cdot y\\(r+s)\cdot x&=r\cdot x+s\cdot x\\(rs)\cdot x&=r\cdot (s\cdot x)\\1\cdot x&=x\end{align}$$

**Remark.** Modules over a field are exactly vector spaces over this field so that the notion of module is a generalisation of vector spaces to rings.

**Observation.** The $\mathbb{Z}$-modules are exactly commutative groups.

**Proof.** The direct implication follows from the definition.

Let $(G,+)$ be a commutative group and let define $\cdot\colon\mathbb{Z}\times G\rightarrow G$ by:
$$0\cdot g:=0_G,n\cdot g:=(n-1)\cdot g+g,(-n)\cdot g:=n\cdot(-g).$$
Then, $(G,+,\cdot)$ is a $\mathbb{Z}$-module. $\Box$

**Observation.** Let $k$ be a field, then $k[T]$-module are exactly the $k$-vector spaces endowed with an endomorphism.

**Proof.** Let $(M,+,\cdot)$ be a $k[T]$-module, then notice that by restriction, $(M,+,\cdot_{\vert k\times M})$ is a $k$-vector space. Furthermore, $\varphi\colon M\rightarrow M$ defined by:
$$\varphi(m):=T\cdot m$$
is an endomorphism of $M$.

Conversely, let $(E,+,\cdot)$ be a vector space and $\varphi\in\textrm{End}(E)$, then let define $\star\colon k[T]\times E\rightarrow E$ by:
$$f\star x:=f(\varphi)(x).$$
Then, $(E,+,\star)$ is a $k[T]$-vector space. $\Box$

**Definition.** Let $M$ be a $R$-module, then $M$ is finitely generated if and only if there exists finetely many elements $x_1,\ldots,x_n$ of $M$ such that:
$$M=\bigoplus_{k=1}^nRx_k:=\left\{\sum_{k=1}^nr_kx_k;r_k\in R\right\}.$$

**Remark.** Respectively, this extends the notion of finitely generated commutative groups and finite-dimensional vector spaces.

**Theorem.** Let $R$ be a principal ideal domain and $M$ be a finitely generated $R$-module, then there exists $d_1\vert\cdots\vert d_n$ elements of $R\setminus R^\times$ such that:
$$M=\bigoplus_{k=1}^nM/(d_k).$$
Furthermore, the $d_k$ are unique up to multiplication by a unit of $R$.

**Proof.** See the corresponding chapter in *Basic Algebra I* by N. Jacobson. $\Box$

**Remark.** This theorem gives the structure of finitely generated commutative group as a direct sum of cyclic groups and the Frobenius decomposition.

Finely, here is the connection I claimed:

In the case of $R=\mathbb{Z}$, the least common multiple of the $d_k$ in the theorem leads to the notion of the exponent of a commutative group and for $R=k[T]$ to the minimal polynomial of an endomorphism.

In a sense, every theorem on the exponent of a commutative group can be transposed in the linear algebra world through the minimal polynomial. Here are a few examples, in particular, you will see how the minimal polynomial offers informations on the linear transformation:

**Proposition.** Let $G$ be a finite abelian group, then $G$ is cyclic if and only if its order equals its exponent.

**Proposition.** Let $E$ be a $n$-dimensional vector space and $\varphi\in\textrm{End}(E)$, then there exists $x\in E$ such that $$\left\{x,\varphi(x),\cdots,\varphi^{n-1}(x)\right\}$$ is a basis of $E$ if and only if the minimal polynomial of $\varphi$ equals its characteristic polynomial.

**Remark.** In the case $R=\mathbb{Z}$, the product of the $d_k$ in the theorem is equal to the order of the group and for $R=k[T]$ to the characteristic polynomial.

**Proposition.** Let $G$ be a group of prime order, then $G$ is a simple group.

**Proposition.** Let $E$ be a $n$-dimensional vector space and $\varphi\in\textrm{End}(E)$, if the minimal polynomial of $\varphi$ is irreducible, then $\{0\}$ and $E$ are the only $\varphi$-invariant subvector spaces of $E$.