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 < 0. Hence 1 − i ∉ H. ❌ Closure fails.

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 or find counterexamples. It cannot prove for all real numbers, but it can demonstrate closure failure.

def in_H(z):
    """Check whether z = a + bi belongs to H."""
    a = z.real
    b = z.imag
    return a * b >= 0

# Counterexample
z1 = complex(2, 1)      # 2 + i
z2 = complex(-1, -2)    # -1 - 2i

print("z1 =", z1, "in H?", in_H(z1))
print("z2 =", z2, "in H?", in_H(z2))

z3 = z1 + z2
print("Sum =", z3, "in H?", in_H(z3))

if not in_H(z3):
    print("Closure fails.")
    

Output

z1 = (2+1j) in H? True
z2 = (-1-2j) in H? True
Sum = (1-1j) in H? False
Closure fails.
    

Extended Program

The following program checks identity, inverses, and closure systematically:

def in_H(z):
    return z.real * z.imag >= 0

identity = complex(0, 0)
print("Identity in H:", in_H(identity))

samples = [complex(2, 1), complex(-3, -2), complex(0, 5), complex(4, 0)]
print("\nInverse Test")
for z in samples:
    print(f"{z} -> {-z} : {in_H(-z)}")

print("\nClosure Counterexample")
z1 = complex(2, 1)
z2 = complex(-1, -2)
print("z1 =", z1, "in H?", in_H(z1))
print("z2 =", z2, "in H?", in_H(z2))
print("Sum =", z1 + z2, "in H?", in_H(z1 + z2))
    

Try It Yourself

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