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!
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?
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number-theory
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2The answer is easy if you are familiar with cyclic groups. Modulo a prime number $p$ you get a complete set of representatives, iff $\gcd(k,p-1)=1$. Are you familiar with primitive roots or cyclic groups? For example, have you seen the fact that if $g$ is a primitive root modulo $p$, then $g^k$ is also primitive,iff $\gcd(k,p-1)$? – 2017-02-04
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0Maybe helpful: your question is equivalent to ask for what exponents $k$ is the function $x\mapsto x^k\mod 11$ a bijection of $\mathbb{Z}_{11}$. – 2017-02-04
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1Alternatively you can get to @Jyrki's conclusion through Fermat's little theorem. (At least in the "if" direction). – 2017-02-04
1 Answers
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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 Hence, $\varphi$ cannot be surjective : for example, $g$ itself doesn't have any antecedent (like any element of order larger than $\lambda$).
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0How about if $\gcd(n,k)\ne 1$? Will $x\mapsto x^k$ necessarily fail to be injective? – 2017-02-04
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1Good question, and this converse should be part of the answer. If $g=\gcd(n,k)$, then the map $x\mapsto x^k$ is the composition of the maps $x\mapsto x^g$ and $x\mapsto x^{k/g}$. And in a finite group of order $n$, if $g\mid n$ then the map $x\mapsto x^g$ is injective if and only if $g=1$. (Consider a prime $p$ dividing $g$ and an element of order $p$.) – 2017-02-04
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0@HenningMakholm: You're right, this property does hold for cyclic groups. I will edit my previous answer. – 2017-02-05