Finite order divides maximal finite order?

Question

Let $G$ be an abelian group, and let $x \in G$ be an element of maximal finite order. If $y \in G$ has finite order in $G$, does it necessarily follow that $|y|$ divides $|x|$?

EDIT: Is there a way to see this without invoking the Fundamental Theorem of Finite Abelian Groups or Lagrange's Theorem...?

EDIT2: Progress so far. We want some sort of contradiction? If $|y|$ is finite and $|y|\,\not\vert\,|x|$, then there exists prime $p$ where $|x| = p^ra$, $|y| = p^sb$, with $\text{gcd}(a, p) = 1$, $\text{gcd}(b, p) = 1$, and $r < s$. But I am not sure what to do next.

What do you have against Lagrange's theorem? – Qudit Sep 09 '15 at 03:00 — Qudit, Sep 09 '15 at 03:00

score 3 · Answer 1 · answered Sep 09 '15 at 07:52

3

Let $x_1=x^{p^r}$, $y_1=y^b$. Then $o(x_1) = a$, $o(y_1)=p^s$, which are coprime, so $o(x_1y_1)=ap^s > ap^r$, contradiction.

answered Sep 09 '15 at 07:52

Derek Holt

96,726

See also http://math.stackexchange.com/questions/1427564/commuting-elements-with-coprime-orders-xy-xy – Cloudscape Sep 13 '15 at 13:57

score 1 · Answer 2 · answered Sep 09 '15 at 02:53

Assuming $G$ is finitely generated, you can appeal to the fundamental theorem and write $G$ as $$\mathbb{Z}^d\oplus \mathbb{Z}/d_1\mathbb{Z} \oplus \mathbb{Z}/d_2\mathbb{Z} \oplus \ldots \oplus \mathbb{Z}/d_k\mathbb{Z}$$ where $d_1\mid d_2, d_2\mid d_3, \ldots, d_{k-1}\mid d_k$.

Now $x$ has order $d_k$ and if $y$ is any other element of finite order, then write $$ y = (y_1,y_2,\ldots, y_k) $$ where $y_i \in \mathbb{Z}/d_i\mathbb{Z}$ and appeal to the fact that $$ |y| = \gcd(|y_1|, |y_2|,\ldots, |y_k|) $$

Finite order divides maximal finite order?

2 Answers2