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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Maths Blog: Subgroups and Centralizers

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Disproving a Subgroup Property

Algorithmic Verification of Algebraic Structures

Exploring Subgroups of Complex Numbers and Centralizers in Dihedral Groups

Abstract

This blog investigates two algebraic problems: verifying whether a constrained subset of complex numbers forms a subgroup under addition, and classifying centralizers in dihedral groups. We show that the set H = {a + bi | a, b ∈ ℝ, ab ≥ 0} fails closure and is not a subgroup of ℂ. We also provide a parity-dependent classification of centralizers in Dn, highlighting differences between odd and even n.

Problem 72: Subgroup of Complex Numbers

Consider H = {a + bi | a, b ∈ ℝ, ab ≥ 0} under addition. To be a subgroup, H must satisfy identity, closure, and inverse properties.

Identity

0 = 0 + 0i ∈ H since 0·0 = 0 ≥ 0. ✅

Closure

Let x = 2 + 0i ∈ H and y = 0 - 3i ∈ H. Then x+y = 2 - 3i. Here ab = 2·(-3) = -6 < 0, so 2 - 3i ∉ H. ❌

Counterexample: x = 2 + 0i, y = 0 - 3i ∈ H, but x+y = 2 - 3i ∉ H.

Conclusion

Since closure fails, H is not a subgroup of ℂ under addition.

Problem 78: Centralizers in Dihedral Group Dn

The dihedral group Dn consists of n rotations and n reflections. For an element g ∈ Dn, the centralizer C(g) = {x ∈ Dn | xg = gx}.

Case 1: n odd

  • C(F) = {e, F} (only identity and the reflection itself)
  • C(R) = {e, R, R², …, Rn-1} (all rotations)

Case 2: n even

  • C(F) = {e, Rn/2, F, F·Rn/2} (identity, half-turn rotation, and two reflections)
  • C(R) = {e, R, R², …, Rn-1} (all rotations)

Summary

  • Odd n: reflections commute only with themselves and identity; rotations commute with all rotations.
  • Even n: reflections through opposite vertices commute with half-turn rotation and related reflections; rotations commute with all rotations.

Conclusion

We proved that H fails subgroup criteria due to lack of closure, and we classified centralizers in Dn based on parity. These results illustrate how continuous and discrete algebraic structures can be rigorously verified and algorithmically formalized.

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