t² | p-1
 

A peaceful and pleasant evening for all,

let p be element of N, t a divisor of p-1
if t² | p-1
is it possible that there exist n^t=1 mod p (n element N)
and it exists no m^(t²) =1 mod p (m element N) at the same time.

Or in other words:
Could there be a cycle subgroup with t elements and at the same time
a non cycle subgroup with t² elements if t² | p-1

If you know an example or a proof for it,
i would be interesting to know it.

Greetings from the primes :geek: :cmd: :truck:
Bernhard

[QUOTE=bhelmes;489571]
is it possible that there exist n^t=1 mod p (n element N)
and it exists no m^(t²) =1 mod p (m element N) at the same time.
[/QUOTE]

What do you think about n=1 and m=1 ?

[QUOTE=bhelmes;489571]
let p be element of N, t a divisor of p-1
if t² | p-1
is it possible that there exist n^t=1 mod p (n element N)
and it exists no m^(t²) =1 mod p (m element N) at the same time.
[/QUOTE]

So not 1 t root of n. Not sure myself at least for non trivial cases.

[QUOTE=bhelmes;489571]A peaceful and pleasant evening for all,

let p be element of N, t a divisor of p-1
if t² | p-1
is it possible that there exist n^t=1 mod p (n element N)
and it exists no m^(t²) =1 mod p (m element N) at the same time.

Or in other words:
Could there be a cycle subgroup with t elements and at the same time
a non cycle subgroup with t² elements if t² | p-1

[snip]
[/QUOTE]
If n^t == 1 (mod p) then n^(t^2) = (n^t)^t == 1 (mod p) as well, so the answer to your question as first stated is "no."

If p is [i]prime[/i], the answer to your question is "no." The multiplicative group is cyclic of order p - 1. If d is any divisor of p - 1, there is at least one element of multiplicative order d.

If you mean, as it appears from your restatement, that "t^2 divides p - 1, there is an element of multiplicative order t, but no element of multiplicative order t^2," then the answer can be "yes," but of course p must be composite.

Example: p = 9, t = 2, n = 8. 8^2 == 1 (mod 9), 2^2 divides 9 - 1 = 8, but there is no element of multiplicative order 4 (mod 9). The multiplicative group (mod 9) is cyclic of order 6.

This example may be generalized. If l is a prime, q is a prime divisor of l - 1, but q^2 is not a divisor of l - 1, then q^2 [i]is[/i] a divisor of l^q - 1. However, the multiplicative group (mod l^q) is cyclic of order l^(q-1)*(l - 1) which is not divisible by q^2. Thus, taking p = l^q, we have that q^2 divides p - 1, there is at least one element of multiplicative order q (mod p), but no element of multiplicative order q^2.

Dear Dr Sardonicus,

thanks for the clear answer. :cmd: :hello::uncwilly:

It was a question which appears to me concerning [url]https://en.wikipedia.org/wiki/Pocklington_primality_test[/url]