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For $k \geq 3$ , the class of $5$ has order $2^{k - 2}$ modulo $2^{k}$

Statement

For every integer $k \geq 3$ , the residue class of $5$ has order $2^{k - 2}$ in $(Z / 2^{k})^{\times}$ .

Facts & Assumptions

Given: An integer $k \geq 3$ .

[L1]

The order is the least positive exponent giving the identity (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity), and an element of order $r$ has $x^{t} = 1$ exactly when $r ∣ t$ (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

[L2]

Group powers satisfy $x^{r + s} = x^{r} x^{s}$ and $(x^{r})^{s} = x^{r s}$ (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

[L3]

$v_{2} (a) = j$ means $2^{j} ∣ a$ and $2^{j + 1} ∤ a$ (For a prime $p$ and a nonzero integer $a$ : $p^{v_{p} (a)} ∣ a$ and $p^{v_{p} (a) + 1} ∤ a$ ; $p^{k} ∣ a$ holds exactly for $k \leq v_{p} (a)$ ; $v_{p} (a) \geq 1$ exactly when $p ∣ a$ ; $v_{p} (1) = v_{p} (- 1) = 0$ ; and $v_{p} (p) = 1$ ), and valuations add on nonzero products ( $v_{p} (a b) = v_{p} (a) + v_{p} (b)$ for nonzero integers $a, b$ , and $v_{p} (a + b) \geq \min {v_{p} (a), v_{p} (b)}$ whenever $a$ , $b$ and $a + b$ are all nonzero).

[L4]

Mathematical induction holds on $N$ (The principle of mathematical induction).

[L5]

A residue class is a unit exactly when its representative is coprime to the modulus (For $n \geq 1$ , $[a]_{n}$ is a unit if and only if $\gcd (a, n) = 1$ ).

Proof

technique · induction

1.1baseL3L5algebra

The odd integer $5$ is coprime to $2^{k}$ , so [L5] puts its class in the unit group. At $j = 0$ , $5^{2^{0}} - 1 = 4$ , so its $2$ -adic valuation is $2$ .

1.2ihL2L3

Assume $v_{2} (5^{2^{j}} - 1) = j + 2$ . Since $5^{2^{j}} \equiv 1 (m o d 4)$ , the factor $5^{2^{j}} + 1$ is congruent to $2$ modulo $4$ and has valuation $1$ . The factorisation $5^{2^{j + 1}} - 1 = (5^{2^{j}} - 1) (5^{2^{j}} + 1)$ and [L3] therefore give valuation $j + 3$ .

2.1step 1.1step 1.2L4

By induction, $v_{2} (5^{2^{j}} - 1) = j + 2$ for all $j \geq 0$ .

3.1step 2.1L3

Step 2.1 with $j = k - 2$ gives $5^{2^{k - 2}} \equiv 1 (m o d 2^{k})$ , while the case $j = k - 3$ gives $5^{2^{k - 3}} \equiv̸ 1 (m o d 2^{k})$ .

4.1step 1.1step 3.1L1discharge-induction∎

By [L1], the order divides $2^{k - 2}$ ; every proper divisor of this prime power divides $2^{k - 3}$ , which step 3.1 excludes. Hence the order is $2^{k - 2}$ .

Depends on

Used by

For k≥3, (ℤ/2ᵏℤ)^×≅ C₂× C_2ᵏ⁻², generated uniquely as (-1)^ε5ʲ Theorem

Dependency tree · two levels

48 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Peter Hackman, Elementary Number Theory, Theorem C.IV.8 (standard reference, not scraped)

For k≥3k≥3, the class of 55 has order 2k−22k−2 modulo 2k2k