Let fG rightarrow H be a homomorphism of groups and let kf

Let f:G rightarrow H be a homomorphism of groups and let k_f = {a epsilon G|f(a) = e_H}, that is, the set of elements of G that are mapped by f to the identity element of H. Prove that k_f is a subgroup of G. See Exercises 34 and 35 for examples.

Solution

let the image f(e_G) has the property

f(e_G) = f(e_{G .} e_G) = f(e_G) . f(e_G)

Left multiplying by f(e_G)^-1 on both sides

f(e_G)^-1.f(e_G) = f(e_{G .} e_G) .f(e_G)^-1

by simplifying we get

e_H= (f(e_G)^-1.f(e_G)).f(e_G) = e_Hf(e_G)=f(e_G)

so the identity in G is mapped to the identity in H.

To check that the image of an inverse is the inverse of an image, compute

f(g^-1).f(g)=f(g^-1.g) = f(e_G) = e_H

using the fact just proven that the identity in G is mapped to the identity in H

Now prove that the kernel is a subgroup of G. The identity lies in the kernel since, as we just saw, it is mapped to

the identity. If g is in the kernel, then g^-1  is also, since, as just showed,

f(g^-1)=f(g)^-1 Finally, suppose both x, y are in the kernel of f. Then

f(xy) = f(x) · f(y) = e_H · e_H = e_H

_{hence proved}