How order of an element and gcd related for abelian groups

Question

How to prove that for any $x \in G$ where $G$ is finite abelian group, $\text{Ord}(x)= \text{Ord}(x^i).\text{gcd}(\text{odd}(x),i)$, $\forall i=1,2,...,\text{ord}(x)$

if $x$ is identity then it is true.

Answer 1

Let $\text{Ord}(x^i) = \text{Ord}(x)/i,\forall i=1,2,...,\text{ord}(x)$

For simplicity let $\text{Ord}(x) = \text{n}.$ i is any divisor of n.

Since $(x^i)^\frac{n}{i} = x^n = e$ then $\text{Ord}(x^i)\le \frac{n}{i}$.

On the other hand, $\forall j\in (0,n/i)$, we have $j.i\le n$

Then $(a^i)^j\neq e$

This implies that $\text{Ord}(x^i)\ge \frac{n}{i}$

Therefore $\text{Ord}(x^i) = \frac{n}{i}$

Since $\text{gcd}(\text{ord}(x^i),i)$ is a divisor of n then the result follows.