how to prove Algebraic structure of the perfect shuffle

Question

二分完美洗牌排列 将$i \rightarrow 2i( mod (2n+1))$.

当$2n+1=3^k$时，很容易得出2分完美洗牌中的$k$个循环结构体,首先我们需要以下定理：

定理1 当p为奇素数且$g$是$p^2$的原根时$\Rightarrow$ $g$也是$p^k$的原根

定义1 欧拉函数，$\phi(m)$定义为小于等于$m$中，与$m$互素的正整数个数

定义2 对模指数，$a$对模$m$的对模指数定义为 $Ord_m(a)=d, st: a^d\equiv1(mod m)$中最小的d

定义3 若$\phi(m)=Ord_m(a)$ 则称$a$为$m$的原根

性质1 若$(a,m)=1$ 且$a$为$m$的原根，那么$a$是$(\Bbb Z/n\Bbb Z)^*$的生成元

例如n=4，则$2n+1=3^2$, ${1,2,3,4,5,6,7,8,9}包含3个循环结构{1,2,4,5,7,8} {3,6}以及{9}

与9互素的正整数集合为{1,2,4,5,7,8}，因此$\phi(m)=6$, 而满足$2^d\equiv1 (mod9)$的最小$d=6$ 因此2为9的原根，同时结合原根的性质2也是整数模9乘法群${1,2,4,5,7,8}$的生成元

考虑包含$3^k$因子的其他循环结构，这些循环结构的元素必定是$3^s2^t$形式 //问题1 如何证明？

当s=0时，循环结构为{1,2,4,5,7,8}

当s=1时，循环结构为{3,6}

当s=2时，循环结构为{9}

问题2，这些结构并非循环群，为何仍然使用生成元，是否正确？

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consider the 2 way perfect shuffle with n=4, that is with elements 1 to 8, how to prove that there are exactly 3 cyclic structures within the shuffle array 1,2,3,4,5,6,7,8. and since not all of the 3 structure are cyclic groups is it correct to use 'generator' to indicating the first element of 3 structures ?

2 way perfect shuffle will map $i \rightarrow 2i( mod (2n+1))$.

when $2n+1=3^k$, it's easy to derive the $k$ cyclic structures within 2 way perfect shuffle permutation. but the conclusion is based on this:

Thorem1 When p is odd prime and $g$ is the primitive root of $p^2$, we also have $\Rightarrow$ $g$ is also the primitive root of $p^k$

Def1 Euler function，$\phi(m)$ is defined as: for positive integers less than $m$，the number count which are co-prime with $m$

Def2 ord_m function，is defined as $Ord_m(a)=d, st: a^d\equiv1(mod m)$ minimum positive integer d

Def3 if $\phi(m)=Ord_m(a)$ then we call $a$ as primitive root of $m$

Property if $(a,m)=1$ and $a$ is the primitive root of $m$，Then $a$ is also the generator for cyclic group $(\Bbb Z/n\Bbb Z)^*$

for example when n=4，$2n+1=3^2$, ${1,2,3,4,5,6,7,8,9}$ contian 3 cyclic structures {1,2,4,5,7,8} {3,6} and {9}

the positive integer colletection coprime with 9 is {1,2,4,5,7,8}，so we have $\phi(m)=6$, while $2^d\equiv1 (mod9)$ the minimum $d=6$ so 2 is primitive root of modulo 9，using the property of primitive root we can derive that 2 is the generator of cyclic group ${1,2,4,5,7,8}$

considering other strucutres that contain element of form $3^k$, these element must be of form $3^s2^t$ //questions1: why ?

when s=0, the structure is {1,2,4,5,7,8}

when s=1, the structure is {3,6}

when s=2, the structure is {9}

question2: can we call 1,3,9 generators of these cyclic structures as not all of them are cyclic groups, and how to better explain the cyclic structure whithin the perfect shuffle.

Answer 1

I'm assuming you understand the theory.

The 'hardest part' of proving this, is to show that $2$ is a primitive root of $\mathbb{Z}_{3^k} ^*$.

Since $\phi(3^k) = 2 \times 3^{k-1}$, this follows if we can show that $2 ^{3^{k-1} } \not\equiv 1 \pmod{3^k} $ and $2^{2 \times 3^{k-1} } \not \equiv 1 \pmod{3^k}$.

The first statement is easy. Expand $2$ as $3-1$, we get

$$ 2^{3^{k-1}} \equiv (3-1) ^ { 3^{k-1} } \equiv -1 \pmod{3^k} $$

The second statement is easy. Expand $2$ as $3-1$, we get

$$ 2^{2\times 3^{k-2}} \equiv (3-1) ^ { 3^{k-1} } \equiv -2\times 3^{k-1} + 1 \pmod{3^k} $$

Hence, $2$ is a primitive root. The cyclic nature of the shuffles are as Jyrki indicated.

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Claim: The numbers that have the form $3^l \times r$, where $ 3 \not \mid r $ would form a cycle.

Proof: This follows directly from $2$ being a primitive root of $3^{k-l}$. Moves consist of going from $a$ to $2a\pmod{3^k}$.
From $3^l \times r$, we can move to all other $3^l \times r^*$, where $3 \not \mid r^*$.
It is also clear that we cannot move to any $3^{l^*} \times r^*$ for $l^* \neq l$.