(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. Suppose that a homomorphism φ:G→ G' is onto. Then its image is all of G', so |imφ|=6. By the First Isomorphism Theorem, G/kerφcong G'; therefore |G|/|kerφ|=|G'|=6. Since |G|=10, this would require |kerφ|=10/6, which is not an integer. But kerφ is a subgroup of G, so by Lagrange’s theorem its order must be an integer divisor of 10. Equivalently, the image of a finite group under a homomorphism has order dividing the order of the original group, while an image inside G' has order dividing 6; hence the image order would have to divide gcd(10,6)=2. It cannot be 6. Thus no onto homomorphism from a group of order 10 to a group of order 6 exists. The conclusion depends only on the orders, not on the particular structures of G and G'.

(b) Principal ideal. In the integral domain Z, every ideal is principal, and the sum of two principal ideals is again principal: aZ+bZ=gcd(a,b)Z. Here 4Z+6Z is the set of all integers of the form 4m+6n, where m,ninZ. Since 4m+6n=2(2m+3n), every such integer is even, so 4Z+6Z⊆ 2Z. Conversely, the greatest common divisor 2 can be written as an integer combination of 4 and 6, namely 2=4(-1)+6(1). Hence 2∈4Z+6Z. If 2k is any even integer, then 2k=k·2, so 2k∈4Z+6Z. Thus 2Z⊆4Z+6Z. Combining the two inclusions gives 4Z+6Z=2Z. Therefore the required principal ideal is 2Z, generated by 2. This also shows that the sum ideal is generated by all integer combinations of the original generators.

(c) Convergence. Let aₙ=(1·3·5…(2n-1))/(2·4·6…(2n))·(x²ⁿ⁺¹)/(2n+1), x>0. The product factor can be written as ((2n)!)/(2²ⁿ(n!)²), but the ratio test is simpler. We have (aₙ₊₁)/(aₙ)=(2n+1)/(2n+2)· x²·(2n+1)/(2n+3)=x²((2n+1)²)/((2n+2)(2n+3)). As n→∞, this ratio tends to x². Hence, by the ratio test, the series converges absolutely when x²<1, i.e. when 0<x<1, and diverges when x²>1, i.e. when x>1. Since all terms are positive for x>0, absolute convergence and ordinary convergence coincide in the convergent cases. The remaining case is x=1, where the ratio tends to 1 and the ratio test is inconclusive. For x=1, apply Raabe’s test. Now (aₙ)/(aₙ₊₁)=((2n+2)(2n+3))/((2n+1)²)=1+(6n+5)/(4n²+4n+1). Therefore n((aₙ)/(aₙ₊₁)-1)=(n(6n+5))/(4n²+4n+1)→frac32>1. Raabe’s test gives convergence at x=1. This is also clear from the asymptotic form of the product factor, which is about 1/√(π n); then the nth term is of order 1/n³/2, a convergent p-series. Consequently, the given series converges for 0<x≤1 and diverges for x>1.

(d) Analyticity of log z. A standard sufficient condition for f(z)=u(x,y)+iv(x,y) to be analytic in a domain D is that u and v have continuous first partial derivatives throughout D and satisfy the Cauchy-Riemann equations uₓ=v_y, u_y=-vₓ at every point of D. For the logarithm, one must choose a single-valued branch. Take the principal branch on the domain D consisting of the complex plane with the non-positive real axis removed. Write z=reⁱθ, r>0, where θ=Argz and -π<θ<π. Then u=log r=frac12log(x²+y²), v=θ. In polar coordinates the Cauchy-Riemann equations are uᵣ=frac1r v_θ and frac1r u_θ=-vᵣ. Here uᵣ=1/r, u_θ=0, vᵣ=0, v_θ=1, so both equations are satisfied. The partial derivatives are continuous on D because r>0 and the chosen argument is smooth away from the branch cut. In Cartesian coordinates the same verification gives uₓ=x/(x²+y²), u_y=y/(x²+y²), vₓ=-y/(x²+y²), v_y=x/(x²+y²), so uₓ=v_y and u_y=-vₓ. Hence log z is analytic on its branch domain. The derivative is df/dz=uₓ+ivₓ=x/(x²+y²)-iy/(x²+y²)=(x-iy)/(x²+y²)=frac1x+iy=frac1z. The same argument works for any branch of log z on a simply connected domain not containing 0.

(e) Linear programming. Let x be the number of jars of P and y the number of jars of Q. The requirements are 2x+2y≥24, 4x+y≥24, x+5y≥20, x,y≥0. The first inequality simplifies to x+y≥12. The cost function to be minimized is Z=30x+50y. The feasible region is the intersection of the half-planes determined by these inequalities. Its relevant corner points are obtained from the boundary lines. The intersection of x+y=12 and 4x+y=24 is (4,8), and it satisfies x+5y=44≥20. The intersection of x+y=12 and x+5y=20 is (10,2), and it satisfies 4x+y=42≥24. The intersection of 4x+y=24 and x+5y=20 is (100/19,56/19), but it is not feasible because x+y=156/19<12. The feasible points on the axes are (0,24) and (20,0). Evaluating the objective function at these vertices gives Z(0,24)=1200, Z(4,8)=520, Z(10,2)=400, Z(20,0)=600. The least value is 400, attained at x=10 and y=2. Thus the person should buy 10 jars of P and 2 jars of Q. This gives A=2(10)+2(2)=24, B=4(10)+1(2)=42, C=1(10)+5(2)=20, meeting all requirements, with minimum cost ₹400. Because the optimal point has integer coordinates, it also satisfies the natural whole-jar interpretation.