[r-t] FCH groups and asymmetric methods

[Philip Saddleton]

Don Morrison said  on 05/03/2007 16:14:
> I believe (though am perhaps mistaken) that the partitioning of false
> course heads into the usual groups (B, D, a, X, etc) depends both upon
> them having Plain Bob lead ends and lead heads (though there are
> isomorphisms to other sets of lead heads) and upon the method having
> the usual palindromic symmetry. What happens when you still have Plain
> Bob lead ends and lead heads, but don't have the usual symmetry.
> Examples of such methods include Selenium Surprise Major and Eastern
> Bob Major. Do the FCHs partition into more or fewer "groups" which
> always appear together? Do these bear some relationship those of the
> usual symmetric methods? Or something else entirely? Or am I mistaken
> and the symmetry is irrelevant?
>   
Corresponding rows in opposite halves of the lead are of opposite 
nature. The groups split into their in and out of course components.