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

[Richard Pullin]

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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