Number of group homomorphisms from $Z_3$ to $Z_6$

Question

So, if $Z_3=\{1,x,x^2\}$ and $Z_6=\{1,x,x^2,x^3,x^4,x^5\}$, and group homomorphisms preserve group structure, then elements from $Z_3$ can only map to 3 distinct elements from $Z_6$ and thus must map to a normal subgroup...

I'm thinking that the identity $1_{Z_3}$ must map to the identity $1_{Z_6}$. This leaves two possibilities for $f(x)$.

If $f(x)=x^2,$ then $f(x^2)=f(x)f(x)=x^2\cdot x^2=x^4$

If $f(x)=x^4$, then $f(x^2)=f(x)f(x)=x^4\cdot x^4=x^8=x^2$

I think these are the only two group homomorphisms. Is this correct?

Answer 1

You claim that "This leaves two possibilities for $f(x)$", but in fact it leaves three possibilities.

As Arthur notes in the comments, group homomorphisms need not be injective. You have given two injective homomorphisms. There is also one non-injective homomorphism.

Answer 2

Number of group homomorphisms from $Z_m$ to $Z_n$ is $=gcd(m,n)$ Just take the generator of$ Z_m$ and try to map it to any element whose order divides $m$ Just try to think that since the generator is the seed of the group every homomorphism is dictated by it.because map any other element is actually mapping the generator's power to something.so why not map the generator only