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:

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

\( |h| \) 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:

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

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:

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

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:

\[ |h^s g^{m^p} | = rp^n \]

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:

\[ |h^s| = p^n \land |g^{m^p}| = s \implies |h^s g^{m^p} | = rp^n \]

Check Aluffi p49 (exercise 1.14) for an explation.

Source

Aluffi p49