[r-t] Hudson stuff

[Richard Smith]

Philip Saddleton wrote:

> Richard Smith wrote:
> >I don't know exactly what you mean by "worthwhile", but all of the
> >larger subgroups of S_6 (upto conjugation)  can be generated by
> >changes:
>
> But what about S_7? The groups of order 21 (clearly impossible) and 168?

(For the benefit of anyone confused by the notation, <1,4,7>
means the group generated by the changes 1, 4 and 7, and
[7.01] is the name given to a group in Brian Price's "The
Composition of Peals in Part".  The "m" or "p" after the
group's order simply states whether it is a mixed parity
group, or a positive parity group.)

For the seven transitive groups, we have:

[7.01] = <1,4,7>     Order 5040m
[7.02] = <23,4,56>   Order 2520p
[7.03]   n/a         Order  168p
[7.04]   n/a         Order   42m
[7.05]   n/a         Order   21p
[7.06] = <1,7>       Order   14m
[7.07]   n/a         Order    7p

Of these, [7.05] and [7.07] are clearly impossible as they
have no elements of order 2.

So nothing new, there.  Looks like seven bells is decidedly
uninteresting in this regard.

Originally, I had restricted myself to the transitive groups
because I had reasoned that any intransitive group could be
expressed as some subdirect product of transitive groups.
I had then assumed that the intransitve group could be
generated by changes if and only if the transitive groups
making it up could be.  Whilst this is clearly necessary,
thinking further about it, it doesn't seem obvious that it
is, in general, sufficient.  Equally I can't think of an
obvious counter example.  Can anyone else?

Richard