Group theory question
Nathan Shirley
Consider the set P(p)={ f(x)=mx+n | m,n in Z_p and m=/=0} where p is a
prime number and f : Z_p-->Z_p.
I would like to find all elements of order p in P(p) by showing P(p) is a
solvable group.
Can anyone offer some help?
I'm not sure what series of normal subgroups with abelian quotients would do
the job for the solvability and then I'm not sure how to use that
information to obtain elements of order p. Thank you.
Arturo Magidin
> Hello.
> Consider the set P(p)={ f(x)=mx+n | m,n in Z_p and m=/=0} where p is a
> prime number and f : Z_p-->Z_p.
> I would like to find all elements of order p in P(p) by showing P(p) is a
> solvable group.
(1) What is the group structure you are giving P(p)? Composition?
Pointwise addition?
(2) How would showing that it is a solvable group help you in finding
all elements of order p, exactly?
> Can anyone offer some help?
> I'm not sure what series of normal subgroups with abelian quotients would do
> the job for the solvability and then I'm not sure how to use that
> information to obtain elements of order p.
So... you would "like" to solve problem X by doing Y, but you have no
idea whatsoever how to use Y to get an answer for X. My question would
be: then why would you say you'd "like" to it in a way for which you
are so clueless? (For that matter, I confess to being utterly clueless
how or why solvability would have anything to do with finding the
elements of order p...)
> Thank you.
You first need to figure out *what* group you are dealing with. Is it
the group of affine functions under composition? Second if the reason
you say you would "like" to do it via proving it is solvable is
because this is a homework assignment, then say so. Otherwise, say why
on Earth you would want to try that particular avenue.
--
Arturo Magidin
Nathan Shirley
Thanks for replying. The problem was given as an exercise during lecture
and I assumed that showing it was solvable would provide a way to find
elements of order p as we were told it was a solvable group (otherwise, why
mention it?). Clearly, I assumed wrong. In any case, I could not see why
it was solvable but I the series 1 < {f(x)=mx} < P(p) shows it is the case.
Yes, the structure on the group is composition, so indeed it is the group of
affine functions under composition. Any suggestions on finding elements of
order p? Thanks.
Arturo Magidin
Sure: find a formula for f composed with f. Use it fo find a formula
for f^p; figure out conditions on n so the constant term becomes 0,
and on m so that the leading term becomes 1. This will give you the
elements of exponent p.
For example, if p=3, then you have
f(f(x)) = f(mx+n) = m(mx+n) + n = m^2 x + (m+1)n
f(f(f(x))) = m(m^2x + (m+1)n) + n = m^3x + (m^2 + m + 1)n.
For f^3(x) = x to hold, you need m^3 = 1 (mod 3) and (m^2+m+1)n = 0
(mod 3). Since x^3 = x (mod 3) for all x (Fermat's Little Theorem),
then m must take a value in {...} and therefore, we get the following
pairs of values for m and n: {...}. Now, throwing out the identity we
have...
Now do it in general.
--
Arturo Magidin
Nathan Shirley
So, in general, f^p(x)=(m^p)*x+(m^(p-1)+m^(p-2)+...+m+1)*n
So we need, m^p=1 mod p, (m^(p-1)+m^(p-2)+...+m+1)*n=0 mod p
By Fermat, m=1 mod p and thus we have p*n=0 mod p and since p=0 mod p, n can
take any value in {0,...,p}
Thus, f(x)=x+n where n in Z_p are all the order p elements.
We have shown f^p=e (e identity) and does it hold by construction that the
order of these elements is nothing smaller than p?
Arturo Magidin
No; if n=0 and m=1, then you get the identity.
In any case, by Lagrange's Theorem you know the order of any of these
elements is a divisor of p. p is a prime. What are the possibilities?
When does an element have order 1?
--
Arturo Magidin
Nathan Shirley
Christmas.