Math Help - Prove [TEX] a^k \neq 1[/TEX] if [TEX]k \mid n[/TEX] and [TEX]k<n[TEX]

1. primitive roots problem

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

So suppose n is an integer and $a \in \mathbb{F}$ and suppose $a^n = 1$.
also $0 < k and $k \mid n$. Prove that if $0 then $a^k \neq 1$.

So far my proof:

Let $a^k =1$, then $gcd(k,n) = d$.
I write the linear combination $d = sk + tn$ for some integers t,n.
then $a^d = 1$

where do i go from here?

2. Re: primitive roots problem

Originally Posted by strider2
also $0 < k and $k \mid n$
Can you clarify this line? Is it meant to say there exists a $k$ such that $k|n$?

3. Re: primitive roots problem

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

-Dan

4. Re: primitive roots problem

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

So suppose n is an integer and $a \in \mathbb{F}$ and suppose $a^n = 1$.
also $0 < k and $k \mid n$. Prove that if $0 then $a^k \neq 1$.

So far my proof:

Let $a^k =1$, then $gcd(k,n) = d$.
I write the linear combination $d = sk + tn$ for some integers t,n.
then $a^d = 1$

where do i go from here?
There is something wrong with the problem statement as I understand it.

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

We have
$a^4 = e$

We have
0 < 2 < 4, 2|4

and
$a^2 = e$

-Dan