Is $m$ divisable by the order of $g$?

Question

Let $G$ be a finite abelian group.

Let $m=\operatorname{Max}\{\textrm{the order of }a:a\in G\}$.

$\forall g\in G$, is $m$ divisable by the order of $g$?

Answer 1

Yes, it's true.

If $G$ is a finite abelian group, then $G$ is isomorphic to a direct sum of the form $$ \mathbb{Z}_{k_1} \oplus \cdots \oplus \mathbb{Z}_{k_n} $$ where $k_i{\,\mid\,}k_{i+1}$, for $1 \le i < n$.

For a reference, see

https://en.wikipedia.org/wiki/Finitely_generated_abelian_group#Invariant_factor_decomposition

It follows that the maximum order for an element of $G$ is $k_n$, and moreover, the order of any element of $G$ is a divisor of $k_n$.

Explanation:

Since $k_i{\,\mid\,}k_{i+1}$, for $1 \le i < n$, it follows that $k_i{\,\mid\,}k_n$, for $1 \le i \le n$.

Assume the operation on $G$ is addition.

For $1 \le i \le n$, let $x_i$ be a cyclic generator of $\mathbb{Z}_{k_i}$.

Then each $x_i$ has order $k_i$, hence, since $k_ix_i=0$, and $k_i{\,\mid\,}k_n$, it follows that $k_nx_i=0$.

Let $y\in G$. Then we can write $$y=w_1x_1 + \cdots + w_nx_n$$ where $w_1,...,w_n\in\mathbb{Z}$. \begin{align*} \text{Then}\;\;k_ny &=k_n(w_1x_1 + \cdots + w_nx_n)\\[4pt] &=w_1(k_nx_1) + \cdots + w_n(k_nx_n)\\[4pt] &=w_1(0) + \cdots + w_n(0)\\[4pt] &=0 \end{align*} Since $k_ny=0$, it follows that the order of $y$ divides $k_n$.