Determine if the given mapping phi is a homomorphism on the given groups. If so, identify its kernel and whether or not the mapping is 1-1 or onto. (a) G is a group, phi: G rightarrow G is defined by phi (a) = a^-1 for a elementof G. (b) G is an abelian group, n > 1 is a fixed integer, and phi: G rightarrow G is defined by phi (a) = a^n for a elementof G. (c) phi: C^x rightarrow R^x with phi (a) = |a|. (d) phi: R rightarrow C^x with phi (x) = cos x + i sin x.

Answer:(a) No. (b)Yes. (c)Yes. (d)Yes.Step-by-step explanation:(a) If [tex]\phi: G \longrightarrow G[/tex] is an homomorphism, then it must hold that [tex]b^{-1}a^{-1}=(ab)^{-1}=\phi(ab)=\phi(a)\phi(b)=a^{-1}b^{-1}[/tex],but the last statement is true if and only if G is abelian.(b) Since G is abelian, it holds that [tex]\phi(a)\phi(b)=a^nb^n=(ab)^{n}=\phi(ab)[/tex]which tells us that [tex]\phi[/tex] is a homorphism. The kernel of [tex]\phi[/tex] is the set of elements g in G such that [tex]g^{n}=1[/tex]. However, [tex]\phi[/tex] is not necessarily 1-1 or onto, if [tex]G=\mathbb{Z}_6[/tex] and n=3, we have [tex]kern(\phi)=\{0,2,4\} \quad \text{and} \quad\\\\Im(\phi)=\{0,3\}[/tex](c) If [tex]z_1,z_2 \in \mathbb{C}^{\times}[/tex] remeber that[tex]|z_1 \cdot z_2|=|z_1|\cdot|z_2|[/tex], which tells us that [tex]\phi[/tex] is a homomorphism. In this case [tex]kern(\phi)=\{\quad z\in\mathbb{C} \quad | \quad |z|=1 \}[/tex], if we write a complex number as [tex]z=x+iy[/tex], then [tex]|z|=x^2+y^2[/tex], which tells us that [tex]kern(\phi)[/tex] is the unit circle. Moreover, since [tex]kern(\phi) \neq \{1\}[/tex] the mapping is not 1-1, also if we take a negative real number, it is not in the image of [tex]\phi[/tex], which tells us that [tex]\phi[/tex] is not surjective.(d) Remember that [tex]e^{ix}=\cos(x)+i\sin(x)[/tex], using this, it holds that[tex]\phi(x+y)=e^{i(x+y)}=e^{ix}e^{iy}=\phi(x)\phi(x)[/tex]which tells us that [tex]\phi[/tex] is a homomorphism. By computing we see that  [tex]kern(\phi)=\{2 \pi n| \quad n \in \mathbb{Z} \}[/tex] and [tex]Im(\phi)[/tex] is the unit circle, hence [tex]\phi[/tex] is neither injective nor surjective.