[r-t] Link between Stedman sixes and Kent fours

[Philip Saddleton]

Any two permutations {a, b} of order two generate a group isomorphic to
a dihedral group, i.e. the symmetries of a regular polygon. The
elements of the group are {a, ab, aba, ..., (ab)^n=i}. If a and b are
changes, there are thus two types of division containing all the
elements of a coset, analogous to Stedman sixes, depending on whether
the first change is a or b. The transposition between division head and
division end is b or a respectively. If c is the change between a pair
of divisions, then substituting aca for bcb in two blocks will allow
them to be joined without introducing additional rows, providing acabcb
has order two. Since the transpositions aca and bcb both swap the same
number of pairs as c, in practice this means the pairs swapping in the
two transpositions satisfy one or more of the following:

- the same pair swapping in each
- two pairs swapping in each, containing the same four bells between
them
- a pair swapping in one but both of these fixed in the other

This is the case for both Stedman sixes and Kent fours, regardless of
the number of bells: adding more bells just means additional pairs swap
in each. Many more examples could easily be constructed.

PABS



On Tue, 2026-09-01 at 16:08 +0100, Richard Pullin wrote:
> A plain course of Stedman Doubles is made up of alternating quick and
> slow sixes, the resulting plain course being the alternating group
> 'A5'. Extents in whole courses are therefore very easy to come by.
> 
> A plain course of Kidderminster Minor is made up of alternating Kent
> and Oxford fours (i.e: the first four rows of Kent or Oxford Treble
> Bob). The resulting plain course of 48 rows is the 'mirror group' as
> ringers call it. Again, extents can be easily constructed out of
> whole courses.
> 
> A plain course of Erin Doubles - slow sixes only - is half of the
> alternating group, with the rows not forming any kind of set. There
> are no possible 120s in whole courses. A plain course of Forward
> Minor - Kent fours only - is half of the mirror group, with the rows
> not forming any kind of set. 720s are possible but not trivial to
> construct and certainly none exist in whole courses.
> 
> Two consecutive slow sixes can be replaced by two consecutive quick
> sixes and v.v. The substitution cleanly joins two blocks together
> without introducing any new rows (e.g: Artistic Triples). The exact
> same scenario is true when substituting two consecutive Kent fours
> with two Oxford fours or v.v (e.g: the Worcester Variation.) I
> struggle to think of other examples where two consecutive blocks of
> changes can be inverted to cleanly join two round blocks together. 
> 
> Does all this demonstrate a group theoretical link going on, perhaps
> something to do with the outer automorphism of S6 as described in
> Brian Price's paper? Or is it just coincidence? Probably it's
> something far more mundane and obvious which I'm just overlooking.
> 
> 
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