[r-t] Conjugate permutations
Richard Smith
richard at ex-parrot.com
Wed Aug 25 10:27:20 UTC 2010
Richard Smith wrote:
> Bizarrely, this function, c(r), turns out to have some very interesting
> properties. [...]
>
> However, what is particularly useful is that it provides an efficient
> algorithm for constructing r from a and b when solving the equation:
>
> b = r^-1 a r.
>
> [...] which gives the solution:
>
> r = c(a) c(b)^-1.
Some people use exponentiation notation to denote
conjugation -- in other words, they define
a^r = r^-1 a r
One inspiration being
(a^r)^s = s^-1 r^-1 a r s = (r s)^-1 a (r s) = a^(rs)
exactly as for real exponents.
Now, I'm a little uncomfortable with this notation because
it seems to invite further extension. For example, if n is
an integer and a and r permutations, then (a^n)^r = (a^r)^n
and it's tempting to write this a^(nr). But what is nr?
But reservations aside, the equation b = r^-1 a r would now
be written:
b = a^r.
If a, b and r were real numbers, we would use logarithms to
solve this for r:
log b = r log a => r = log b / log a.
(I've written log rather than ln as the choice of base is
irrelevant.)
For permutations, the function c(r)^-1 serves the same role:
r = c(a) c(b)^-1 => c(b)^-1 = c(a)^-1 r.
If you're going to use exponentiation to represent
conjugation, is it unreasonable to define
log r = c(r)^-1 ?
I'm not sure how much more mileage there is in this, though.
RAS
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