Let $a \in \mathbb{F}$. Prove that if $0 < k < n$, then $a^k \neq 1$ Let $n$ be an integer with $n > 0$, Let $a \in  \mathbb{F}$, and suppose that $a^n = 1$.
Also suppose that if $0 < k < n$ and $k \mid n$,
then $a^k \neq 1$. Prove that if 
$0 < k < n$, then $a^k \neq 1$.
Can anyone give me a clue in which to solve the previous problem?
 A: Assume by contradiction that $a^k=1$ for some $1<k<n$. Let $d=\gcd(k,n)$. Then $d<n$.
Write $d$ as a linear combination of $k,n$ and prove that $a^d=1$. But this is a contradiction (why?)....
A: Expanded Proof:
Since we've already assumed that $a^k\ne1$ for $0<k<n$ with $k\mid n,$ then it remains only to show that $a^k\neq 1$ for $0<k<n$ with $k\not\mid n$.
Suppose $0<k<n$ with $k\not\mid n$ and let $d=\gcd(k,n)$. Now, since $\Bbb Z$ is a Bézout domain, then there exist $x,y\in\Bbb Z$ such that $d=kx+ny$, so if we did have $a^k=1$, then we would consequently have $$\begin{align}a^d &= a^{kx+ny}\\ &= a^{kx}a^{ny}\\ &= (a^k)^x(a^n)^y\\ &= 1^x1^y\\ &= 1.\end{align}$$ But $0<d<n$ and $d\mid n,$ so this is impossible. Thus, $a^k\ne1,$ as desired.
A: Hint $\rm\ A^N\! = 1 = A^K \Rightarrow\: A^{(N,K)}\! = 1\:$ by Bezout (or, since the set of $\rm\:K\:$ such that $\rm\:A^K = 1\:$ is closed under subtraction, so closed under gcd).
A: Phrased slightly differently: we have a map of groups $$\mathbb{Z} \rightarrow \mathbb{F}^*$$ given by $1 \mapsto a$. The kernel is of the form $d\mathbb{Z}$. Thus taking $d > 0$, we have $0 < d \leqslant n$, $d|n$. Your condition says then that $d = n$, as desired. 
