Math and science::Algebra::Aluffi

Groups. All orders divides order of maximal element

Order: divides order of maximal element

Let $G$ be a commutative group, and let $g \in G$ be an element with maximal finite order. Any other element with finite order has an order that divides $| g |$ .

The same statement cannot be made for non-comutative groups.

Can you remember the proof?

Maximal finite order

Let $g$ be an element of group $G$ . $g$ has maximal finite order iff all other elements with finite order have an order less than that of $g$ .

Proof essence

Consider elements $g$ and $h$ of group $G$ , with orders as follows:

\begin{aligned} | g | & = 2^{4} 3^{7} 5^{3} \\ | h | & = 2^{4} 3^{8} 5^{1} \end{aligned}

$| h |$ doesn't divide $| g |$ , as one prime is raised to a larger value. Commutativity allows us to create a new element $f$ that has an order whose prime factors are the maximum of those from $| g |$ and $| h |$ . This order must be larger than $| g |$ , so $g$ can't have maximal order.

In comparison, if $k$ is another element and $| k | = 2^{4} 3^{5} 5^{1}$ , then this order divides $| g |$ , as all primes are raised to a smaller power in $| k |$ .

Proof

Proof.

Let $g$ and $h$ be elements of a group $G$ , with $g$ being an element of maximal finite order. Proceed by contradiction: assume that $| h |$ does not divide $| g |$ .

We can express $| g |$ and $| h |$ in terms of their prime factorization:

\begin{aligned} | g | & = 2^{a} 3^{b} 5^{c} 7^{d} . . . \\ | h | & = 2^{i} 3^{j} 5^{k} 7^{l} . . . \end{aligned}

If every prime of $| h |$ is raised to a lower corresponding power than $| g |$ , then $| h |$ would divide $| g |$ . But by assumption it does not, so there must be at least 1 prime that is raised to a higher power in $| h |$ compared to $| g |$ . Denote this prime by $p$ . Focusing on this prime, we can now express $| g |$ and $| h |$ as so:

\begin{aligned} | g | & = p^{m} r \\ | h | & = p^{n} s \end{aligned}

Where $r$ and $s$ collect the other primes.

In a controlled way, we can create elements with smaller orders than $| g |$ and $| h |$ . Consider first $| g |$ . The element $g^{m}$ has an order that is $m$ times smaller than $| g |$ , $| g^{m} | = \frac{| g |}{m}$ . This is so, as $m$ is a factor of $| g |$ . Complete this reduction by considering the element $g^{m^{p}}$ , which is an element with order $| g^{m^{p}} | = r$ . We have found an element whose order is $| g |$ but with the prime $p$ absent.

The same process can be repeated for $| h |$ , and we now consider the element $h^{s}$ , which has order $| h^{s} | = p^{n}$ .

When we combine these two elements into a third element, the point of this process becomes clear. The element $h^{s} g^{m^{p}}$ has an order:

| h^{s} g^{m^{p}} | = r p^{n}

As $n > m$ , we have found an element with order larger than that of $| g |$ .

Importance of commutativity

The proof above doesn't specifically mention where commutativity comes in. Can you spot it? Possibly surprisingly, implication requiring commutativity is the statement:

| h^{s} | = p^{n} \land | g^{m^{p}} | = s ⟹ | h^{s} g^{m^{p}} | = r p^{n}

Check Aluffi p49 (exercise 1.14) for an explation.

Source

Aluffi p49