Answer:
(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.