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Verify whether the following mapping is a homomorphism. If so, determine its kernel.

$f(x)=x^3$, for all $x$ belonging to $G$.
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I think definition of group G should  be given in the question ie. set and its associated operation.

and it should also be mentioned that mapping is from group G to same group G or other group.

I think, question is split up into 2 parts. Part of this question is given in this link :- https://gateoverflow.in/94630/gate1988-13ia

So, here, it is given that G is the group of non zero real numbers under multiplication.

As we know that , In homomorphism ,

if mapping f : (G,*) ---> (H,*) for groups G and H

then   f(a*b) = f(a)*f(b) for all a,b ∈ G

I assumed that mapping is :- f : (G,*) ---> (G,*)

So, let , y and z are the 2 elements of G such that  y = p/q and z = r/s and p,q,r,s all are non-zero integers because G is a set of non-zero real numbers.

Since , here mapping is defined as f(x) = x3

So,      f(a*b) = f(a)*f(b)

f(y*z) = f(y)*f(z)

f((p/q)*(r/s)) = f(p/q)*f(r/s)

(pr/qs)3 = (p/q)3 * (r/s)3

(pr/qs)3 = (pr/qs)3

So, we can say that given mapping is Homomorphism.

According to definition of kernel ,

ker f = {g ∈ G : f(g) = eH

Since , identity element for multiplication operation on non-zero real numbers is 1

so , eH = 1

and f(1) = 13 = 1

So, g = 1

So, ker f = 1

https://en.wikipedia.org/wiki/Homomorphism

https://en.wikipedia.org/wiki/Kernel_(algebra)