(a) Prove that a non-commutative group of order 2p, where p is an odd prime, must have a subgroup of order p. (15 marks) (b)…

(a) Subgroup of order p in a non-commutative group of order 2p

Let |G| = 2p, where p is an odd prime. By Lagrange’s theorem, every element of G has order 1, 2, p or 2p. Since G is non-commutative, it cannot be cyclic; hence no element can have order 2p, because an element of order 2p would generate a cyclic group of order 2p, which is abelian.

Suppose G has no element of order p. Then every non-identity element must have order 2. For any a, b ∈ G, we then have a² = b² = e and also (ab)² = e. Thus abab = e, so ab = b⁻¹a⁻¹ = ba, since a⁻¹ = a and b⁻¹ = b. Hence G would be abelian, a contradiction.

Therefore G must contain an element a of order p (the only remaining possibility). Then H = ⟨a⟩ is a subgroup of G and |H| = p. Thus a non-commutative group of order 2p must have a subgroup of order p.

(b) Minimum and maximum distances by Lagrange multipliers

Let Q(x, y, z) be any point on the sphere x² + y² + z² = 4. The square of the distance from P(2, 6, 3) to Q is

S = (x − 2)² + (y − 6)² + (z − 3)².

We minimize and maximize S subject to the constraint g = x² + y² + z² − 4 = 0. By Lagrange’s multiplier method, at an extremum,

∇S = λ∇g.

Thus

2(x − 2) = 2λx, 2(y − 6) = 2λy, 2(z − 3) = 2λz.

So

(1 − λ)x = 2, (1 − λ)y = 6, (1 − λ)z = 3.

Let μ = 1 − λ. Then

x = 2/μ, y = 6/μ, z = 3/μ.

Using x² + y² + z² = 4,

(4 + 36 + 9)/μ² = 4,

so 49/μ² = 4, giving μ² = 49/4, hence μ = ±7/2.

For μ = 7/2, the point is (4/7, 12/7, 6/7). The distance squared is

S = (4/7 − 2)² + (12/7 − 6)² + (6/7 − 3)² = (−10/7)² + (−30/7)² + (−15/7)² = (100 + 900 + 225)/49 = 1225/49 = 25.

So the minimum distance is √25 = 5.

For μ = −7/2, the point is (−4/7, −12/7, −6/7). The distance squared is

S = (−4/7 − 2)² + (−12/7 − 6)² + (−6/7 − 3)² = (−18/7)² + (−54/7)² + (−27/7)² = (324 + 2916 + 729)/49 = 3969/49 = 81.

So the maximum distance is √81 = 9.

Minimum distance = 5 at (4/7, 12/7, 6/7); maximum distance = 9 at (−4/7, −12/7, −6/7).

(c) The integral by contour integration

Put z = e^iθ. Then

cos θ = (z + z⁻¹)/2, cos 2θ = (z² + z⁻²)/2, dθ = dz/(iz).

On the unit circle |z| = 1,

I = ∮ over |z|=1 [cos 2θ/(5 + 4 cos θ)] dθ = ∮ [((z² + z⁻²)/2)/(5 + 4((z + z⁻¹)/2))] dz/(iz) = ∮ (z⁴ + 1)/(2i z²(2z² + 5z + 2)) dz.

Now

2z² + 5z + 2 = (2z + 1)(z + 2).

Hence

f(z) = (z⁴ + 1)/(2i z²(2z + 1)(z + 2)).

The poles inside |z| = 1 are z = 0, of order 2, and z = −1/2, a simple pole. The pole z = −2 lies outside the unit circle.

Residue at z = −1/2:

Res(f, −1/2) = [(−1/2)⁴ + 1]/[2i(−1/2)²(−1/2 + 2)·2] = (17/16)/[2i·(1/4)·(3/2)·2] = (17/16)/(3i/2) = −17i/24.

Residue at z = 0: write

f(z) = (z⁴ + 1)/(2i z²(2 + 5z + 2z²)).

Since

1/(2 + 5z + 2z²) = 1/2 − 5z/4 + ⋯,

we get

f(z) = ((z⁴ + 1)/(2i z²))(1/2 − 5z/4 + ⋯) = (1 + z⁴)/(2i z²)(1/2 − 5z/4 + ⋯) = 1/(4i z²) − 5/(8i z) + ⋯.

Thus the residue at z = 0 is

−5/(8i) = 5i/8.

Sum of residues inside |z| = 1 is

5i/8 − 17i/24 = (15i − 17i)/24 = −i/12.

By Cauchy’s residue theorem,

I = 2πi × (−i/12) = π/6.

∫ from 0 to 2π of cos 2θ/(5 + 4 cos θ) dθ = π/6.