# [r-t] algebraic theory

Martin Bright martin at boojum.org.uk
Thu Jan 31 11:04:11 UTC 2013

```On 31 January 2013 12:48, Robin Woolley <robin at robinw.org.uk> wrote:
> I believe that given any two elements p and q, then the orders of p and q
> divide the order of the (finite) group. (A brief scan of the literature
> fails to explcitly confirm this).

Yes: the order of an element p is the order of the cyclic subgroup it
generates, which by Lagrange's Theorem divides the order of the parent
group.

> Question: Is there an easy way of deciding the maximum order of an element
> in a finite group - I believe that for S5 this equals 6?

I can easily believe that there's no good algorithm for this in
general, for a group specified by finitely many generators and
relations.  However, for permutation groups (i.e., groups explicitly
given as subgroups of Sn) it's pretty straightforward: any permutation
can be written as a product of disjoint cycles, and it's then easy to
see that the order is the lcm of the lengths of the cycles.  So, in
S5, if you take a permutation with a 2-cycle and a 3-cycle (say,
21534) then the order is 6.  No other combination of cycle lengths can
do any better.

> Question: Since pq is in the group also, what is the order of pq? (i.e., if
> ord(p) = 2 and ord(q) = 4, then is it easy to write down ord(pq)? - I don't
> think it can be eight, for instance).

No, there can be no good rule for this.  For example, if p has order
27, and q is the inverse of p, then q also has order 27 but pq is the
identity and has order 1.  On the other hand, it's easy to write down
a group with two elements p and q each of order 2 but where pq has
order as large as you like.  What is true (and easy to prove) is that,
if p and q commute and have coprime orders, then the order of pq is
the product of the orders of p and q.

Algebra lesson over.

Martin

```

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