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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Disproving a Subgroup Property

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Disproving a Subgroup Property

Disproving the Property for Subgroups

Suppose H is a nonempty subset of a group G with the property:

If a, b ∈ H, then a-1b-1 ∈ H.

Is this enough to guarantee H is a subgroup?

The answer is no. While every subgroup satisfies this property, a subset can satisfy it without being a subgroup, typically by failing to include the identity element.

Subgroup Criteria

  • Non-empty
  • Closed under the group operation
  • Contains the identity element
  • Closed under inverses

First Counterexample

Consider the cyclic group Z3 = {0,1,2} under addition mod 3.

Let H = {1}.

  • Non-empty: Yes, H contains 1.
  • Property check: For a = b = 1, we compute (-1) + (-1) = -2 ≡ 1 (mod 3). Since 1 ∈ H, the property holds.
  • Subgroup check: H does not contain the identity 0. Also, 1+1 = 2 ∉ H, and the inverse of 1 is 2 ∉ H. Therefore, H is not a subgroup.

Second Counterexample (General Case)

Let a be an element of order 3 in any group G (so a3 = e). Consider the singleton set H = {a}.

  • Non-empty: H contains a.
  • Property check: For a = b = a, we have a-1a-1 = (a2)-1 = (a-1)-1 = a ∈ H. So the property holds.
  • Subgroup check: H does not contain the identity e (since a ≠ e). Also, H is not closed under the group operation: a·a = a2 ∉ H. Therefore, H is not a subgroup.

Conclusion

The property a, b ∈ H ⇒ a-1b-1 ∈ H is not sufficient to guarantee H is a subgroup. For subgroup status, we must additionally require that H contains the identity element (or an equivalent condition ensuring closure under the group operation).

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