Subgroup Verification in Complex Numbers

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Subgroup Verification in Complex Numbers Subgroup Verification in Complex Numbers Testing subgroup properties of H = {a + bi ∈ ℂ ∣ ab ≥ 0} Mathematical Solution Define H = {a + bi ∈ ℂ ∣ a, b ∈ ℝ, ab ≥ 0} . That is, the real and imaginary parts must have the same sign (or one of them is zero). 1. Identity The additive identity in ℂ is 0 + 0i. Since 0·0 = 0 ≥ 0, we have 0 ∈ H. ✅ 2. Closure Take z₁ = 2 + i and z₂ = −1 − 2i. Both satisfy ab ≥ 0. Their sum is 1 − i, and 1×(−1) = −1 3. Inverse For z = a + bi ∈ H, we have ab ≥ 0. Its inverse is −z = −a − bi. Then (−a)(−b) = ab ≥ 0, so −z ∈ H. ✅ Conclusion ✔ Identity exists ✔ Inverses exist ✘ Closure fails Therefore, H is not a subgroup of (ℂ, +). Python Verification A Python program can test many examples to provide evidence ...
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Algorithmic Verification of Algebraic Structures

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Algorithmic Verification of Algebraic Structures

Algorithmic Verification of Algebraic Structures

Subgroups of Complex Numbers and Centralizers in Dihedral Groups

Abstract

The formal verification of algebraic properties within continuous and discrete mathematical frameworks represents a critical intersection of abstract algebra and computational theory. This paper investigates the structural integrity of specific algebraic sets, focusing on a constrained subset of complex numbers and the symmetric properties of dihedral groups. Through rigorous analytical deduction, we evaluate the subgroup criteria for the set H = {a + bi | a, b ∈ ℝ, ab ≥ 0} under addition, demonstrating its failure to satisfy the closure property. Furthermore, we systematically classify the centralizers for reflection and rotation elements within the dihedral group Dn, delineating the divergent algebraic behaviors based on the parity of n.

Introduction

Group theory provides the foundational language for understanding symmetries and structural invariants across mathematics and physics. The complex plane ℂ serves as a rich topological space where subsets can be evaluated for group-theoretic properties such as closure, identity, and invertibility. Simultaneously, dihedral groups Dn offer a discrete environment to study the rigid symmetries of regular n-sided polygons, encapsulating both rotations and reflections.

This study defines its scope around two primary algebraic challenges: the verification of a specific subset of ℂ as a subgroup (Problem 72), and the determination of centralizers within Dn (Problem 78).

Method/Approach

Verification Framework for Complex Subgroups

For Problem 72, the target set is H = {a + bi | a, b ∈ ℝ, ab ≥ 0}. Identity holds, but closure fails. Example: x = 2 + 0i ∈ H, y = 0 - 3i ∈ H, yet x+y = 2 - 3i ∉ H since 2·(-3) = -6 < 0.

Centralizer Computation in Dihedral Groups

For Problem 78, rotations commute with all rotations, so C(R) = {e, R, R², …, Rn-1}. - If n is odd: C(F) = {e, F}. - If n is even: C(F) = {e, Rn/2, F, F·Rn/2}.

Discussion

These results highlight the importance of counterexample generation in continuous sets and parity-based classification in discrete groups. They also show how algorithmic verification can unify continuous and discrete algebraic reasoning.

Conclusion

We demonstrated that H fails to form a subgroup of ℂ under addition due to closure violation, and we classified centralizers in Dn based on parity. Bridging these rigorous proofs with algorithmic frameworks advances automated theorem verification.

References

  • Malestein & Theran (2013). Frameworks with forced symmetry I. Link
  • Kuckert (2004). Spin, Statistics, and Reflections. Link
  • Chiu & Schnyder (2015). Classification of crystalline topological semimetals. Link
  • Shi et al. (2020). SymmetryNet. Link
  • Funk & Liu (2017). Beyond Planar Symmetry. Link
  • Seo et al. (2022). Equivariant Learning. Link
  • Vaz & Mann (2019). Clifford Algebraic Description. Link
  • Buchweitz et al. (2018). The magic square of reflections and rotations. Link

Rotations

For g = rk, we get rks = sr−k. Commuting requires r−k = rk ⇒ r2k = e. Since n is odd, this forces k = 0. Thus only the identity rotation commutes.

Reflections

For g = srk, we compute (srk)s = r−k and s(srk) = rk. Equality requires r−k = rk ⇒ k = 0. Hence only F itself commutes.

Therefore: C(F) = {e, F}, with order 2.

2. Centralizer of a Reflection when n is Even

Rotations

r2k = e ⇒ k = 0 or k = n/2. So two rotations commute: e and rn/2.

Reflections

For g = srk, commuting requires r−k = rk ⇒ k = 0 or k = n/2. So two reflections commute: s and srn/2.

Therefore: C(F) = {e, rn/2, F, Frn/2}, with order 4.

3. Geometric Explanation

Odd n: Each reflection axis passes through one vertex and the midpoint of the opposite edge. No two reflection axes are opposite, so only identity and the reflection itself commute.

Even n: Each reflection axis has an opposite axis. A 180° rotation (rn/2) maps the axis onto itself, so the half-turn and opposite reflection also commute. Hence the centralizer doubles in size.

4. Centralizer of a Non-Identity Rotation

Let R = rk, k ≠ 0.

  • All rotations commute with each other since ⟨r⟩ is cyclic.
  • A reflection commutes only if rk = r−k ⇒ 2k ≡ 0 (mod n).

Therefore:
- If R ≠ rn/2, then C(R) = ⟨r⟩ (all rotations), order n. - If n is even and R = rn/2, then the half-turn is central: C(R) = Dn.

Summary Table

Element n odd n even
Reflection F {e, F} {e, rn/2, F, Frn/2}
Rotation R ≠ e, rn/2 All rotations All rotations
Half-turn rn/2 — Entire Dn 

def dihedral_group(n):
    """
    Elements are represented as:
    ('r', k)  -> rotation r^k
    ('s', k)  -> reflection s r^k
    """
    G = [('r', k) for k in range(n)] + [('s', k) for k in range(n)]
    return G

def multiply(a, b, n):
    t1, k1 = a
    t2, k2 = b

    if t1 == 'r' and t2 == 'r':
        return ('r', (k1 + k2) % n)
    elif t1 == 'r' and t2 == 's':
        return ('s', (k2 - k1) % n)
    elif t1 == 's' and t2 == 'r':
        return ('s', (k1 + k2) % n)
    else:  # ('s','s')
        return ('r', (k2 - k1) % n)

def centralizer(n, element):
    G = dihedral_group(n)
    C = []
    for g in G:
        if multiply(g, element, n) == multiply(element, g, n):
            C.append(g)
    return C

def pretty(x):
    t, k = x
    if t == 'r':
        return "e" if k == 0 else f"r^{k}"
    else:
        return "s" if k == 0 else f"sr^{k}"

# Example usage
n = 15
print(f"Dihedral group D_{n}")

F = ('s', 0)
print("Centralizer of reflection s:", [pretty(x) for x in centralizer(n, F)])

R = ('r', 3)
print("Centralizer of rotation r^3:", [pretty(x) for x in centralizer(n, R)])

if n % 2 == 0:
    H = ('r', n // 2)
    print("Centralizer of half-turn:", [pretty(x) for x in centralizer(n, H)])

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