[r-t] Plain Bob leadheads

[Martin Bright]

Here is a proof.  Probably not the shortest possible.

Let e be the lead end, a the lead-end change and h the Plain Bob lead head.

Claim: We have aha = h^-1.
Proof: You get h by applying a to e, so h = ea.  The method is
palindromic, meaning e = e^-1.  Therefore aha = a(ea)a = ae = (ea)^-1
= h^-1.

Claim: The only changes a satisfying aha=h^-1 are 12 and 1n.
Proof:  It's enough to prove it in the case h = (2 4 6 .. n n-1 ... 5
3) in cycle notation.
Easy to check (how conjugation works in permutation groups): you find
aha by applying a to each number appearing in this cycle, that is,
  aha = (a_2 a_4 a_6 ... a_n a_(n-1) ... a_5 a_3)  where a_i denotes
the ith bell in the row you get by applying a to rounds.
This needs to be equal to h^-1 = (3 5 ... n-1 n ... 6 4 2), so these
sequences of numbers need to be the same up to rotation.
So a has to fix the treble.  Under a, either 2 stays fixed or 2 moves
to 3rds place.
If 2 stays fixed, you need (2 a_4 a_6 ... a_n a_(n-1) ... a_5 a_3) =
(2 3 5 ... n-1 n ... 6 4), so a is the change 12.
If 2 -> 3, then 3 -> 2 and you need (3 a_4 a_6 ... a_n a_(n-1) ... a_5
2) = (3 5 ... n-1 n ... 6 4 2), so a is the change 1n.

Martin

On Fri, 19 Feb 2021 at 13:14, Martin Bright <martin at boojum.org.uk> wrote:
>
> Hi Mark,
>
> These seem to be the only two changes that fix the treble and
> normalise the subgroup of Plain Bob lead heads - at least my computer
> tells me this is true on 8.  I wonder whether that can be turned into
> a proof.
>
> All the best,
> Martin
>
> On Fri, 19 Feb 2021 at 10:15, Mark Davies <mark at snowtiger.net> wrote:
> >
> > Please can someone produce, or point me at, a concise proof of this lemma:
> >
> > PB1: In a palindromic treble-dominated method with Plain Bob leadheads,
> > on an even number of bells, the lead end place notations must be 12 or 1n.
> >
> > Thank you!
> >
> > MBD
> >
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