Re: primitive roots problem

Quote:

Originally Posted by

**strider2** also $\displaystyle 0 < k<n$ and $\displaystyle k \mid n$

Can you clarify this line? Is it meant to say there exists a $\displaystyle k$ such that $\displaystyle k|n$?

Re: primitive roots problem

Another clarification...is $\displaystyle \mathbb{F}$ a group $\displaystyle \mathbb{F}( \cdot )$? So I can presume that 1 is the multiplicative identity?

-Dan

Re: primitive roots problem

Quote:

Originally Posted by

**strider2** I'm trying to figure this problem out and need a little help.

So suppose n is an integer and $\displaystyle a \in \mathbb{F}$ and suppose $\displaystyle a^n = 1$.

also $\displaystyle 0 < k<n$ and $\displaystyle k \mid n$. Prove that if $\displaystyle 0<k<n$ then $\displaystyle a^k \neq 1$.

So far my proof:

Suppose a contradiction.

Let $\displaystyle a^k =1$, then $\displaystyle gcd(k,n) = d$.

I write the linear combination $\displaystyle d = sk + tn$ for some integers t,n.

then $\displaystyle a^d = 1$

where do i go from here?

There is something wrong with the problem statement as I understand it.

Counter-example. Let $\displaystyle \mathbb{F}$ be the multiplicative group of two elements e and a.

We have

$\displaystyle a^4 = e$

We have

0 < 2 < 4, 2|4

and

$\displaystyle a^2 = e$

-Dan