ord$_{2^k}(5)=2^{k-2}$

hbghlyj

Powers of 2

Modulo $4$ there are two coprime congruence classes, $[1]$ and $[3]$, so $ (\mathbb {Z} /4\mathbb {Z} )^{\times }\cong \mathrm {C} _{2}, $ the cyclic group with two elements.

Modulo $8$ there are four coprime congruence classes, $[1]$, $[3]$, $[5]$ and $[7]$. The square of each of these is $1$, so $ (\mathbb {Z} /8\mathbb {Z} )^{\times }\cong \mathrm {C} _{2}\times \mathrm {C} _{2}, $ the Klein four-group.

Modulo $16$ there are eight coprime congruence classes $[1]$, $[3]$, $[5]$, $[7]$, $[9]$, $[11]$, $[13]$ and $[15]$. $ \{\pm 1,\pm 7\}\cong \mathrm {C} _{2}\times \mathrm {C} _{2}, $ is the 2-torsion subgroup (i.e., the square of each element is 1), so $ (\mathbb {Z} /16\mathbb {Z} )^{\times } $ is not cyclic. The powers of $3$, $ \{1,3,9,11\} $ are a subgroup of order 4, as are the powers of $5$, $ \{1,5,9,13\}$. Thus $ (\mathbb {Z} /16\mathbb {Z} )^{\times }\cong \mathrm {C} _{2}\times \mathrm {C} _{4}. $

The pattern shown by $8$ and $16$ holds for higher powers $2^k, k > 2:$ $ \{\pm 1,2^{k-1}\pm 1\}\cong \mathrm {C} _{2}\times \mathrm {C} _{2}, $ is the 2-torsion subgroup (so $ (\mathbb {Z} /2^{k}\mathbb {Z} )^{\times } $ is not cyclic) and the powers of $3$ are a cyclic subgroup of order $2^{k-2}$, so $ (\mathbb {Z} /2^{k}\mathbb {Z} )^{\times }\cong \mathrm {C} _{2}\times \mathrm {C} _{2^{k-2}}. $

$\DeclareMathOperator{\ord}{ord}$Lemma 3.7的证明使用$\ord_{2^k}(5)=2^{k-2}$
可以用 3 代替 5 吧？$\ord_{2^k}(3)=2^{k-2}$
仿照上面的证明$3^{2^{k-3}}=(2^2-1)^{2^{k-3}}$ 二项展开？$=1-2^{k-1}+\sum_{i\ge2}\binom{2^{k-3}}i2^{2i}$

hbghlyj

计算一些小例子，
Mod 8:
$⟨3⟩=⟨5⟩=\{1,3\}$
Mod 16:
$⟨3⟩=\{1,3,9,11\}$
$⟨5⟩=\{1,5,9,13\}$

若$q≠3,5$，$\ord_{2^k}(q)=2^{k-2}$还成立吗？

Czhang271828

hbghlyj 发表于 2023-5-15 23:45
若$q≠3,5$，$\ord_{2^k}(q)=2^{k-2}$还成立吗？

$\mathbb Z_{2^k}^\times \simeq \mathbb Z_2\times \mathbb Z_{2^{k-2}}$, 阶数为 $2^{k-2}$ 的元素一共 $2^{k-2}$ 个. 取值为 $4\cdot\text{奇数}\pm 1$.