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.

(a) No. (b)Yes. (c)Yes. (d)Yes.

Step-by-step explanation:

(a) If is an homomorphism, then it must hold

that ,

but the last statement is true if and only if G is abelian.

(b) Since G is abelian, it holds that

which tells us that is a homorphism. The kernel of

is the set of elements g in G such that . However,

is not necessarily 1-1 or onto, if and

n=3, we have

(c) If remeber that

, which tells us that is a

homomorphism. In this case

, if we write a

complex number as , then , which tells

us that is the unit circle. Moreover, since

the mapping is not 1-1, also if we take a negative

real number, it is not in the image of , which tells us that

is not surjective.

(d) Remember that , using this, it holds that

which tells us that is a homomorphism. By computing we see

that   and

is the unit circle, hence is neither injective nor

surjective.

answer: -10

step-by-step explanation:

s(t) = 1 - 10t

s(x) = 1 - 10x

s(x + h) = 1 - 10(x + h)

            = 1 - 10x - 10h

s(x + h) - s(x) = 1 - 10x - 10h - (1 - 10x)

                      = 1 - 10x - 10h - 1 + 10x

                      = -10h

= -10