Homomorphisms $V_4$ and $\mathbb{Z}/4\mathbb{Z}$ How many homomorphisms exist between the integers modulo $4$ under addition, and the Klein-four-group? Also, how do I find them? Cheers.
 A: The homomorphisms from $\Bbb Z/n\Bbb Z$ to $G$, for an arbitrary group $G$,
correspond to the solutions of $a^n=e$ in $G$. The homomorphism corresponding
to such an $a$ takes $k$ to $a^k$.
So, how many solutions of $a^4=e$ are there in $V_4$?
A: $V_4=\{e,a,b,ab\}$ and $Z_4=\{0,1,2,3 \}$ and let $f:V_4 \to  Z_4$ be a homomorphism then $o(f(a)) \vert o(a)$ then $f(a)=0 ,1$ similarly $f(b)=0,1$ so in this way u may get $4$ homomrphisms from  $V_4 \to Z_4$.
A: Suppose that $f: G \to H$ and $o(a)$ denotes the order of $a$ in $G$. Assume that $G$ and $H$ are finite groups. We can write
$$e_H=f(e_G)=f(a^{o(a)})=(f(a))^{o(a)}$$
This implies that $o(f(a)) \mid o(a)$. In general, if $g^n=e$ in a group and $g$ has finite order, then $o(g) | n$. You can prove this as an exercise but it's a well-known fact in algebra. The proof involves dividing $n$ by $o(g)$ using Euclid's algorithm and showing that $n=o(g)q+r$ implies that $r=0$ because $o(g)$ is the smallest exponent that gives identity.
Now suppose that $f: \mathbb{Z}_4 \to V_4$. We know that $\mathbb{Z}_4$ is cyclic. Hence, it suffices to see where $[1]_4$ is sent under $f$. You can check that there are no order restrictions here because all elements in $V_4$ are of order $2$ and $1$ which divide $4$. Hence, you can send $[1]_4$ to any of the elements $\{e,a,b,ab\}$ and check that you'll get four homomorphisms this way. Although, they look very similar to each other.
For the case when $f: V_4 \to \mathbb{Z}_4$, Sajan has already explained the idea I wrote in the comments wonderfully.
A: There is no epimorphism $f\colon\mathbb{Z_4}\to\mathbb{V_4}$ because $\mathbb{Z_4}$ is cyclic but $\mathbb{V_4}$ is not. There is no epimorphism  $g\colon\mathbb{V_4}\to\mathbb{Z_4}$  because $\circ(\bar1)=4 \nmid\circ(a),\forall a \in\mathbb{V_4}$,because the order of all non-identity elements in $\mathbb{V_4}$ is $2$
Now, let $\phi\colon\mathbb{Z_4}\to\mathbb{V_4}$ be a homomorphism.
The group $\phi(\mathbb{Z_4})$ is a subgroup of $\mathbb{V_4}$,so  $|\phi(\mathbb{Z_4})|$ divides $|\mathbb{V_4}|$ i.e. $|\phi(\mathbb{Z_4})|=1,2,4$ 
$|\phi(\mathbb{Z_4})|\ne4$ because as mentioned above there is no epimorphism.
If $|\phi(\mathbb{Z_4})|=1$,then it is the trivial  homomorphism.
If $|\phi(\mathbb{Z_4})|=2$,then the $\phi(\bar1)=e$ and the other three elements of $\mathbb{Z_4}$ map to the same non-identity element of $\mathbb{V_4}$ and there are three different ways of doing that because there are only three non-identity elements in $\mathbb{V_4}$.
All total $4$ homomorphisms are there from $\mathbb{Z_4}$ to $\mathbb{V_4}$.
