For which exponents k is $\{1^k,2^k,3^k,4^k,5^k,6^k,7^k,8^k,9^k,10^k,11^k\}$ a complete set of representatives modulo 11? For which exponents k is $\{1^k,2^k,3^k,4^k,5^k,6^k,7^k,8^k,9^k,10^k,11^k\}$ a complete set of representatives modulo 11?  I can see that clearly $k=1$ gives a complete set of representatives and $k=2$ gives $\{1, 4, 9, 5, 3, 3, 5, 9, 4, 1, 0\}$ which is not a complete set, but I'm not sure how to proceed.  Also is there a way to generalize an answer to this to other moduli?  Any help is appreciated, thanks!
 A: Given a finite group $(G,.)$ of order $n$ and any integer $k$ such that $gcd(n,k)=1$, the map $\varphi:G\to G,x\mapsto x^k$ is bijective.
Indeed, there exist integers $u,v$ such that $un+vk=1$. For every $x\in G$, we have $x^n=1$, hence $x=x^{un+vk}=(x^v)^k$. Beeing surjective, $\varphi$ is also bijective since $G$ is finite.
Now, for any prime $p$, the set $\left(\mathbb{Z}/p\mathbb{Z}\right)^\star$ is a multiplicative group of order $p-1$. Hence, if $gcd(p,k)=1$, we see that $1^k$, $2^k$, ..., $(p-1)^k$ is a set of representatives mod. $p$. And it remains to add $0^k=0$.
The original question corresponds to the special case $p=11$.
EDIT
Conversely, as mentioned in the comments below, the hypothesis that $\varphi$ is bijective does not imply that $gcd(k,n)=1$.
This is however true for cyclic groups. Here is a proof ...
Consider a cyclic group $G$ of order $n$ and an integer $k$ such that $gcd(n,k)=d>1$. Let $\lambda,\mu$ be positive integers such that :
$$n=\lambda d,\quad k=\mu d,\quad gcd(\lambda,\mu)=1$$
Let $g\in G$ be a generator of $G$. It is known that the order of any element of $G$ can be computed by the formula :
$$ord(g^i)=\frac{n}{gcd(n,i)}$$
As a consequence, we have for every $i\in\{0,\cdots,n-1\}$ :
$$ord(\varphi(g^i))=ord(g^{ki})=\frac{n}{gcd(n,ki)}=\frac{\lambda d}{gcd(\lambda d,\mu di)}=\frac{\lambda}{gcd(\lambda,\mu i)}\le\lambda<n$$
Hence, $\varphi$ cannot be surjective : for example, $g$ itself doesn't have any antecedent (like any element of order larger than $\lambda$).
