Please explain your reasoning in the questions below. Writing it on paper and po

Question

Please explain your reasoning in the questions below.  Writing it on paper and posting photos would be easiest--since typing mathematical symbols may be a bit tedious.  Thank you so much!

Consider the space G = C([0,1]) of real-valued functions which are continuous on [0, 1] (in the usual calculus sense). We make G into a group under ordinary addition of functions: if f,g G, then f + g G is defined by (f + g)(x) = f(x) + g(x). What is the identity of G? What is the inverse of f G? Let H = {f G : f(1/3 ) = f(2/3 ) = 0}. Show that H

Explanation / Answer

a) the zero function is the identity and the inverse of f is -f

indeed f+0=0+f=f and f-f=0

b) suppose f,g are in H . let k a scalar then (f-kg)(1/3)=f(1/3)-kg(1/3)=0 ( same for 2/3)

so H is a subgroup of G. it's obviously a proper subgroup since not all function verifies that (e.g f(x)=x)

c)

H={f/ f(1/3)=f(2/3)=0 }

Let phi : G -> R^2

f-> (f(1/3),f(2/3))

Ker(phi)= H by definition of H.

phi is surjective because if we take any (a,b) in R^2 then we can find a continuous function such that f(1/3)=a and f(2/3)=b. For that we can take the Lagrange interpolation function f(x) = (x-1/3)(3b)- (x-2/3)(3a).

So Im(phi)=R^2

By the first isomorphism theorem G/ker(phi) ~ Im(phi)

Hence you have G/H ~ R^2

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