(a) Let G be a finite group of order mn, where m and n are prime numbers with m n. Show that G has at most one subgroup of order m. 10 marks

(b) If w = f(z) is an analytic function of z, then show that

((∂²)/(∂ x²) + (∂²)/(∂ y²)) log |f'(z)| = 0.

Question

(a) Let G be a finite group of order mn, where m and n are prime numbers with m > n. Show that G has at most one subgroup of order m. 10 marks

(b) If w = f(z) is an analytic function of z, then show that

((∂²)/(∂ x²) + (∂²)/(∂ y²)) log |f'(z)| = 0. 10 marks

(c) Test the convergence of ∫limits₀² (log x)/(√((2-x))) dx . 10 marks

(d) If φ and ψ are functions of x and y satisfying Laplace equation, then show that f(z) = p + iq, i = √−1 is an analytic function, where p = (∂ φ)/(∂ y) - (∂ ψ)/(∂ x) and q = (∂ φ)/(∂ x) + (∂ ψ)/(∂ y) . 10 marks

(e) Use two phase method to solve the following linear programming problem :

Maximize z = x₁ + 2x₂

subject to x₁ - x₂ ≥ 3

2x₁ + x₂ ≤ 10

x₁, x₂ ≥ 0

10 marks

UPSC Answer Check · Accepted Answer

(a) Use Sylow’s third theorem. Since |G| = mn, with m and n primes and m > n, any subgroup of order m is a Sylow m-subgroup. Let s be the number of Sylow m-subgroups of G. Sylow’s theorem gives
s ≡ 1 (mod m), and s divides |G|/m = n.
Since n is prime, the only divisors of n are 1 and n. Thus s = 1 or s = n. If s = n, then n ≡ 1 (mod m), i.e. m divides n − 1. But 1 ≤ n < m, so n − 1 < m, impossible. Hence s = 1.
Therefore G has exactly one, hence at most one, subgroup of order m.
Final answer: G has at most one subgroup of order m.

(b) Let F(z) = f′(z) = u(x, y) + i v(x, y). Since f is analytic, F is also analytic, so u and v satisfy the Cauchy–Riemann equations
u_x = v_y, u_y = −v_x.
Assume F(z) ≠ 0 where the logarithm is taken. Put
L = log |f′(z)| = log |F(z)| = (1/2) log(u² + v²).
Since F is analytic and non-zero, log F(z) is locally analytic. Its real part is exactly log |F(z)|. Write log F = U + i V, where U = log |F|. For an analytic function, the real and imaginary parts satisfy the Cauchy–Riemann equations:
U_x = V_y, U_y = −V_x.
Therefore
U_xx + U_yy = ∂/∂x(V_y) + ∂/∂y(−V_x) = V_yx − V_xy = 0,
because mixed partial derivatives are equal. Hence
(∂²/∂x² + ∂²/∂y²) log |f′(z)| = 0.
Final answer: the Laplacian of log |f′(z)| is zero wherever f′(z) ≠ 0.

(c) Let
I = ∫₀² log x / √(2 − x) dx.
The integrand has possible singularities only at x = 0 and x = 2 because log x → −∞ as x → 0⁺ and 1/√(2 − x) → ∞ as x → 2⁻.

Near x = 2, for 3/2 < x < 2, log x is bounded by log 2, so
|log x| / √(2 − x) ≤ C / √(2 − x).
But ∫₀^ε t^(−1/2) dt converges because 1/2 < 1. Hence x = 2 also causes no divergence.
Thus the integral converges.

For its exact value, put x = 2t. Then dx = 2 dt and √(2 − x) = √2 √(1 − t). Hence
I = √2 ∫₀¹ [log(2t)] / √(1 − t) dt
= √2 [ log 2 ∫₀¹ dt/√(1 − t) + ∫₀¹ log t / √(1 − t) dt ].
Now
∫₀¹ dt/√(1 − t) = 2,
and by the beta function,
∫₀¹ log t / √(1 − t) dt = 4 log 2 − 4.
Therefore
I = √2 [2 log 2 + 4 log 2 − 4] = √2(6 log 2 − 4).
Final answer: the integral converges, and its value is 6√2 log 2 − 4√2.

(d) We are given that φ and ψ satisfy Laplace’s equation:
φ_xx + φ_yy = 0, ψ_xx + ψ_yy = 0.
Define
p = φ_y − ψ_x, q = φ_x + ψ_y.
To prove f(z) = p + i q is analytic, it is enough to verify the Cauchy–Riemann equations p_x = q_y and p_y = −q_x.

First,
p_x = φ_yx − ψ_xx,
q_y = φ_xy + ψ_yy.
Since mixed partial derivatives commute, φ_yx = φ_xy. Also ψ_xx + ψ_yy = 0, so ψ_yy = −ψ_xx. Hence
q_y = φ_xy − ψ_xx = p_x.

Next,
p_y = φ_yy − ψ_xy,
q_x = φ_xx + ψ_yx.
Using φ_xx + φ_yy = 0, we get φ_yy = −φ_xx. Also ψ_xy = ψ_yx. Thus
−q_x = −φ_xx − ψ_yx = φ_yy − ψ_xy = p_y.
So p_y = −q_x.

Thus p and q satisfy the Cauchy–Riemann equations. Assuming φ and ψ are C², p and q have continuous first partial derivatives, so f(z) = p + i q is analytic.
Final answer: f(z) = p + i q is analytic.

(e) Convert the problem to standard form. The constraints are
x₁ − x₂ ≥ 3,
2x₁ + x₂ ≤ 10,
x₁, x₂ ≥ 0.
Introduce surplus s₁, artificial a₁, and slack s₂:
x₁ − x₂ − s₁ + a₁ = 3,
2x₁ + x₂ + s₂ = 10,
x₁, x₂, s₁, s₂, a₁ ≥ 0.

Phase I: Minimize W = a₁.
Initially a₁ = 3, s₂ = 10, and x₁ = x₂ = s₁ = 0. From the first equation,
W = a₁ = 3 − x₁ + x₂ + s₁.
Choose x₁ to enter because increasing x₁ decreases W. The ratio test gives min(3/1, 10/2) = 3, so a₁ leaves. Pivot on x₁:
x₁ − x₂ − s₁ = 3,
3x₂ + 2s₁ + s₂ = 4.
Now W = 0. Since W ≥ 0 always, Phase I is complete.

Phase II: Use the original objective
Maximize z = x₁ + 2x₂.
The current equations are
x₁ − x₂ − s₁ = 3,
3x₂ + 2s₁ + s₂ = 4.
Express z in terms of nonbasic variables:
x₁ = 3 + x₂ + s₁,
so
z = x₁ + 2x₂ = 3 + 3x₂ + s₁.
Enter x₂. The ratio test gives 4/3, so s₂ leaves. Pivot on x₂:
x₂ = 4/3 − (2/3)s₁ − (1/3)s₂.
Then
x₁ = 3 + x₂ + s₁ = 13/3 + (1/3)s₁ − (1/3)s₂.
Substitute into z:
z = x₁ + 2x₂ = 7 − s₁ − s₂.
Since the coefficients of s₁ and s₂ in z are negative, no further increase is possible. Hence the maximum occurs at
x₁ = 13/3, x₂ = 4/3.
Check: x₁ − x₂ = 13/3 − 4/3 = 3, and 2x₁ + x₂ = 26/3 + 4/3 = 10.
Final answer: maximum z = 7 at x₁ = 13/3 and x₂ = 4/3.

(a) Let G be a finite group of order mn, where m and n are prime numbers with m > n. Show that G has at most one subgroup of…

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