How to find all homomorphisms from $(Z_{11}^*,\times)$ to $(Z_6,+)$ The task is to find all homomorphisms from $Z_{11}^*$ to $Z_6$ and other way around.
I start by writing all the elements of these groups and their orders.
$Z_{11}^*$ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and orders are 1, 10, 5, 5, 5, 10, 10, 10, 5, 2.
Z6 = {0, 1, 2, 3, 4, 5} with orders 1, 6, 3, 2, 3, 6.
I start with $Z_{11}^*$ to Z6:
I know that a homomorphism F must satisfy F(1)=0 (neutral element must be mapped to neutral element) and ord(F(n)) | ord(n). 
$Z_{11}^*$ = <2>, $ord_{Z_{11}^*}(2)$=10 and only $ord_{Z_6}(0)$=1 | 10 or $ord_{Z_6}(3)$=2 | 10. 
I don't know if it doesn't matter if I choose one generator from $Z_{11}^*$ and find all possible homomorphism, or I must try every generator from $Z_{11}^*$. 
Suppose that F(2) = 0 holds, so F must be a zero homomorphism. 
Suppose that F(2) = 3 and F(1) = 0 holds, so F(2×2) = F(4)= 3+3 = 6 = 0, F(8) = 3, ... and I obtain F(2i) = 3 if i is odd and 0 if i is even. 
So I have found two homomorphism.
I don't know if it is true that generators must be mapped to generators. If so, I have 2 -> 1 or 2 -> 5 but ord(1)=ord(5)=6 and doesn't divide ord(2)=10. So where 2 souhld be mapped? Or it holds that the homomorphism is given by the image of generator, even when the generator isn't mapped to generator?
Homomorphism F from $Z_6$ to $Z_{11}^*$:
$Z_6$ = <1>, $ord_{Z_6}(1)$ = 6. In $Z_{11}^*$ there are only two elements satisfying ord(F(n)) | ord(n). $ord_{Z_{11}^*}(1)$ = 1 and $ord_{Z_{11}^*}(10)$ = 2.
Suppose that F(0) = 1 and F(1) = 1 holds. This case isn't possible. 
Suppose that F(0) = 1 and F(1) = 10 holds. F(2) = F(1+1) = 10 × 10 = 1. F(3) = 10, ... 
So I obtain F(i) = 1 if i is even, F(i) = 10 if i is odd. 
Is it correct? Have I found all homomorphisms? 
Thank you.
 A: It is correct and you have found all homomorphisms. If group is cyclic, it's enough to choose one generator $f$ and check its image - if there are two homomorphisms $h$ and $g$ s.t. $h(a) = g(a)$, then $h \equiv g$: for any element $x$ in domain we have $x = n \cdot a$ for some integer $n$ (where $n \cdot a = a + a + \ldots + a$ [$n$ times] - it's well-defined due to associativity). And then we have $f(x) = f(n\cdot a) = n\cdot f(a) = n\cdot g(a) = g(n \cdot a) = g(x)$.
In the second part - $F(0) = F(1) = 1$ is possible. It will be just trivial homomorphism (everything to neutral element) - remember that neutral element of $Z_{11}^*$ is $1$, not $0$.
You can also significantly reduce amount of work by using group decomposition. $11$ is prime, so $Z_{11}$ is field, so $Z_{11}^*$ is cyclic. As it has $10$ elements, it's isomorphic to $Z_{10}$ and also to $Z_2 \times Z_5$. $Z_6$ is isomorphic to $Z_2 \times Z_3$. Fix some decompositions, so now we need only find all homomorphisms from $Z_2$ to $Z_2$, $Z_2$ to $Z_3$, $Z_5$ to $Z_2$ and $Z_5$ to $Z_3$ - we can create a homomorphism from $Z_{11}^*$ to $Z_6$ from any set of this smaller homomorphisms, or decompose it.
As $2$, $3$ and $5$ are coprime, the only homomorphisms from $Z_2$ to $Z_3$ and $Z_5$ to $Z_2$ or $Z_3$ are trivial. There are two homomorphisms from $Z_2$ to $Z_2$ - trivial and identity; they correspond to the two homomorphisms you have found.
