$G$ is a finite abelian group and $m:=\max \{o(x):x \in G\}$ , then is it true that $o(x)|m , \forall x \in G$?

Question

If $G$ is a finite abelian group and $m:=\max \{o(x):x \in G\}$ , then is it true that $o(x)|m , \forall x \in G$ ?

Answer 1

Yes. This is an immediate consequence of the structure theorem for finitely generated Abelian groups.

More precisely, write $G = \mathbb{Z}/(a_1) \times \dots \times \mathbb{Z}/(a_n)$, where $a_1 \mid a_2 \mid \dots \mid a_n$. Then $a_n$ is the largest order of an element, but also a multiple of the order of every element.

Answer 2

Yes. Write $G=P_1\times\cdots\times P_n$, where the $P_i$ are the Sylow-$p_i$ subgroups of $G$. Let $x_i$ be the maximum order of any element of $P_i$, with the element of maximal order denoted $g_i$. Then $g=(g_1,\ldots, g_n)$ has order $x_1\cdots x_n$. This is the $m$ we seek, since any $h=(h_1,\ldots, h_n)\in G$ has order equal to the product of the orders of $h_i\in P_i$, each of which is at most $x_i$. Now, in addition, $o(h_i)|o(g_i)$, since each $P_i$ is a $p_i$-group and so all orders are powers of $p$, with $x_i$ having the maximal exponent on $p_i$. Thus, $o(h)|o(g)$ for all $h\in G$.

Answer 3

Yes, it is true, by contradiction: Let $g$ be the element of max order and let $h$ be an element whose order is not a divisor of the order of $g$,then the order of $h$ beats the order of $g$ for some prime $p$. Let $j$ be so that the order of $g^j$ is the order of $g$ divided by the maximum common denominator of the order of $g$ and the order of $g^j$. Let $g^j=k$.

Notice $k$ and $h$ have relatively prime orders, so $\langle k\rangle \cap \langle h \rangle= \{e\}$ by lagrange.

now notice $(kh)^n=k^nh^n=e$ if and only if both are $e$ (since the generated subgroups have trivial intersection). therefore the order of $kh$ is the least common multiple of the order of $k$ and the order of $h$ which is the product of the orders since they are relatively prime. This contradicts the maximality of the order of $g$ since the order of $h$ is larger than the order of the gcd of the orders of $g$ and $h$.

Answer 4

Here is an elementary way to prove this:

First, we note that it will suffice to prove the following:
For any $x,y\in G$ there is a $z\in G$ with $o(z) = \mathrm{lcm}(o(x),o(y))$.

We then prove it in the special case where $\mathrm{gcd}(o(x),o(y)) = 1$ by noting that in this case, we have $o(xy) = o(x)o(y)$.

Finally, we reduce to this case by noting that if we let $n = \mathrm{gcd}(o(x),o(y))$ and let $x' = x^n$ then $o(x') = \frac{o(x)}{\mathrm{gcd}(o(x),o(y))}$, so $\mathrm{gcd}(o(x'),o(y)) = 1$ and $\mathrm{lcm}(o(x),o(y)) = o(x')o(y)$ (i.e. the element $z$ we are looking for is $x'y$).