Answer:
(a) No. (b)Yes. (c)Yes. (d)Yes.
Step-by-step explanation:
(a) If  is an homomorphism, then it must hold
 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 a homorphism. The kernel of  
 
is the set of elements g in G such that  . However,
. However, 
 is not necessarily 1-1 or onto, if
 is not necessarily 1-1 or onto, if  and
 and 
n=3, we have 

(c)  If  remeber that
 remeber that
 , which tells us that
, which tells us that  is a
 is a 
homomorphism. In this case 
 , if we write a
, if we write a 
complex number as  , then
, then  , which tells
, which tells 
us that  is the unit circle. Moreover, since
 is the unit circle. Moreover, since 
 the mapping is not 1-1, also if we take a negative
 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
, which tells us that 
 is not surjective.
 is not surjective.
(d) Remember that  , using this, it holds that
, using this, it holds that

which tells us that  is a homomorphism. By computing we see
 is a homomorphism. By computing we see 
that   and
 and 
 is the unit circle, hence
 is the unit circle, hence  is neither injective nor
 is neither injective nor 
surjective.