Is Cubing a Permutation in Zp for Primes Equivalent to 2 Mod 3?

[Hello Kitty]

[SOLVED] cubic reciprocity?

I would like to prove the following conjecture:

If $$p \equiv 2\ (mod\ 3)$$ is a prime, then the cubing function $$x \mapsto x^3$$ is a permutation of $$\mathbb{Z}_p$$.

I've tried to find a contradiction to the negation by assuming that if $$n \neq m\ (mod\ 3)$$, but $$n^3 \equiv m^3\ (mod\ 3)$$, then since $$m^3 - n^3 = (m-n)(m^2-mn+n^2)$$, we must have $$(m^2-mn+n^2) \equiv 0\ (mod\ 3)$$ to avoid a contradiction concerning zero-divisors. Now I'm stuck.

 

[al-mahed]

what exactly you mean saying "the cubing function $$x \mapsto x^3$$ is a permutation of $$\mathbb{Z}_p$$"?

We know that there are $$2^p$$ permutations (sub-groups) in $$\mathbb{Z}_p$$, right?

So you mean that one of these subgroups represent a cube? If yes, is the sub-group sum or product?

 

[HallsofIvy]

I would have interpreted "the cubing function $$x \mapsto x^3$$is a permutation of $$\mathbb{Z}_p$$" to mean that it permutes the members of $$\mathbb{Z}_p$$- but that's so trivial, that's probably not what's meant!

0³= 0, 1³= 1, 2³= 2 mod 3. In fact, the cubing function is just the identity on $$\mathbb{Z}_p$$!

 

[Hurkyl]

3 isn't the only prime.


As for the original poster... what do you know about the multiplicative structure of Z/pZ?

 

[Hello Kitty]

I thought "the cubing function $$x \mapsto x^3$$ is a permutation of $$\mathbb{Z}_p$$" was a pretty unambiguous statement. Maybe it would be better to say it's a "bijection $$\mathbb{Z}_p \rightarrow \mathbb{Z}_p$$".

HallsofIvy: Besides the fact that 3 is not congruent to 2 mod 3, in $$\mathbb{Z}_5$$, $$2^3 \equiv 3\ (mod\ 5)$$ so it is not the identity.

Hurkyl: It is a group of order p-1, which is of the form 3k+1. I guess the obvious fact that springs to mind is that no element can have order 3 then by Lagrange. This means no cube can be congruent to 1 mod p (except, of course, 1). So I guess this means the kernel of the cubing map is {1} ***alarm bells ringing*** actually this map is a group homomorphism so Ker={1} implies it is injective. Of course this implies it is a bijective too.

 

[robert Ihnot]

It can be looked at a little more algebratically: If X^3 == Y^3 Mod P, then there exists an element x/y such that (x/y)^3 == 1 Mod P, which implies 3 divides the order of the group, p-1. Thus, unless x=y, P would be of the form 3k+1. (Which is about what Hello Kitty is saying)