(a) Let G be a group of order 10 and G′ be a group of order 6. Examine whether there exists a homomorphism of G onto G′. (10…

(a) Homomorphism from a group of order 10 onto a group of order 6

No such homomorphism exists. Suppose φ : G → G′ is a homomorphism onto G′, with |G| = 10 and |G′| = 6. Its kernel K = ker φ is a subgroup of G, and by the First Isomorphism Theorem G/K ≅ G′. Hence

|G|/|K| = |G′|, so |K| = 10/6,

which is not an integer. Equivalently, by Lagrange's theorem |K| must be a divisor of 10, namely 1, 2, 5 or 10, and these give images of order 10, 5, 2 and 1. None equals 6.

Therefore there is no homomorphism of G onto G′. The conclusion uses only the orders and holds whatever the structures of G and G′.

(b) The ideal 4ℤ + 6ℤ as a principal ideal

The ideal 4ℤ + 6ℤ is the set of all integers 4m + 6n with m, n ∈ ℤ.

Since 4m + 6n = 2(2m + 3n), every element is even, so 4ℤ + 6ℤ ⊆ 2ℤ.

Conversely, 2 = 4(−1) + 6(1) lies in 4ℤ + 6ℤ, so every multiple 2k = k·2 also lies in it. Hence 2ℤ ⊆ 4ℤ + 6ℤ.

Therefore 4ℤ + 6ℤ = 2ℤ = ⟨2⟩, a principal ideal generated by gcd(4, 6) = 2. In general aℤ + bℤ = gcd(a, b)ℤ, because ℤ is a principal ideal domain.

(c) Convergence of the series

Let aₙ = [1·3·5⋯(2n−1)/(2·4·6⋯(2n))] · x²ⁿ⁺¹/(2n+1), with x > 0. All terms are positive.

Ratio test: aₙ₊₁/aₙ = [(2n+1)/(2n+2)] · x² · [(2n+1)/(2n+3)] → x² as n → ∞.

So the series converges if x² < 1, that is 0 < x < 1, and diverges if x² > 1, that is x > 1.

For x = 1 the ratio test fails, so use Raabe's test: aₙ/aₙ₊₁ = (2n+2)(2n+3)/(2n+1)² = 1 + (6n+5)/(4n²+4n+1), n(aₙ/aₙ₊₁ − 1) = n(6n+5)/(4n²+4n+1) → 3/2 > 1.

Hence the series converges at x = 1. (This agrees with the estimate 1·3⋯(2n−1)/(2·4⋯2n) ≈ 1/√(πn), which makes the nth term of order n^(−3/2).)

The series converges for 0 < x ≤ 1 and diverges for x > 1.

(d) Analyticity of log z

Sufficient conditions: f(z) = u(x, y) + iv(x, y) is analytic in a domain D if u and v and their first partial derivatives u_x, u_y, v_x, v_y are continuous in D and satisfy the Cauchy-Riemann equations u_x = v_y, u_y = −v_x at every point of D.

For log z take a single-valued branch, the principal branch on D = ℂ minus the non-positive real axis. With z = re^(iθ), r > 0, −π < θ < π, u = log r = (1/2) log(x² + y²), v = θ = arg z = arctan(y/x) (suitably continued).

Then u_x = x/(x²+y²), u_y = y/(x²+y²), v_x = −y/(x²+y²), v_y = x/(x²+y²).

These are continuous in D, and u_x = v_y and u_y = −v_x. So log z is analytic in D (and likewise for any branch on a simply connected domain that excludes 0).

Derivative: df/dz = u_x + i v_x = (x − iy)/(x²+y²) = 1/(x + iy) = 1/z.

(e) Minimum-cost purchase of jars

Let x be the number of jars of P and y the number of jars of Q. Minimise Z = 30x + 50y subject to 2x + 2y ≥ 24 (chemical A), that is x + y ≥ 12, 4x + y ≥ 24 (chemical B), x + 5y ≥ 20 (chemical C), x, y ≥ 0.

The feasible region is unbounded, lying above all three lines. Its corner points are:

• On the y-axis, (0, 24) where 4x + y = 24 meets x = 0. It satisfies x + y ≥ 12 and x + 5y ≥ 20.
• (4, 8), where x + y = 12 meets 4x + y = 24.
• (10, 2), where x + y = 12 meets x + 5y = 20.
• On the x-axis, (20, 0) where x + 5y = 20 meets y = 0. It satisfies the other two.

The intersection of 4x + y = 24 and x + 5y = 20, namely (100/19, 56/19), has x + y = 156/19 < 12, so it is not feasible.

Values of Z: Z(0, 24) = 1200, Z(4, 8) = 520, Z(10, 2) = 400, Z(20, 0) = 600.

The least value is 400 at (10, 2). Because the region is unbounded and the cost coefficients are positive, this minimum is attained.

The person should buy 10 jars of P and 2 jars of Q, at the minimum cost of ₹400. Check: A = 2(10) + 2(2) = 24, B = 4(10) + 2 = 42, C = 10 + 5(2) = 20, so all requirements are met.