Vtyjkyuk

How many homomorphisms exist between the integers modulo $4$ under addition, and the Klein-four-group? Also, how do I find them? Cheers.

$V_4=e,a,b,ab$ 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 $V_4 to Z_4$.

The homomorphisms from $Bbb Z/nBbb Z$ 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 $V_4$?

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 $f: mathbbZ_4 to V_4$. We know that $mathbbZ_4$ 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 $f: V_4 to mathbbZ_4$, Sajan has already explained the idea I wrote in the comments wonderfully.

There is no epimorphism $fcolonmathbbZ_4tomathbbV_4$ because $mathbbZ_4$ is cyclic but $mathbbV_4$ is not. There is no epimorphism $gcolonmathbbV_4tomathbbZ_4$ because $circ(bar1)=4 nmidcirc(a),forall a inmathbbV_4$,because the order of all non-identity elements in $mathbbV_4$ is $2$

Now, let $phicolonmathbbZ_4tomathbbV_4$ be a homomorphism.

The group $phi(mathbbZ_4)$ is a subgroup of $mathbbV_4$,so $|phi(mathbbZ_4)|$ divides $|mathbbV_4|$ i.e. $|phi(mathbbZ_4)|=1,2,4$

$|phi(mathbbZ_4)|ne4$ because as mentioned above there is no epimorphism.

If $|phi(mathbbZ_4)|=1$,then it is the trivial homomorphism.

If $|phi(mathbbZ_4)|=2$,then the $phi(bar1)=e$ and the other three elements of $mathbbZ_4$ map to the same non-identity element of $mathbbV_4$ and there are three different ways of doing that because there are only three non-identity elements in $mathbbV_4$.

All total $4$ homomorphisms are there from $mathbbZ_4$ to $mathbbV_4$.