# How to determine the total number of homomorphisms from $V_4 \rightarrow C_2$?

How to determine the total number of homomorphisms from $V_4 \rightarrow C_2$? Where $V_4$ is the Klein-$4$ group and $C_2$ is cyclic of order $2$.

So far I have managed to create $7$ just from playing about with the elements.

These are: The trivial homomorphism. $3$ homomorphisms in which all elements of $V_4$ are mapped to the identity in $C_2$ except one non identity element which is mapped to the non identity in $C_2$. And finally the $3$ homomorphisms taking $2$ distinct non-identity elements from $V_4$ and mapping them to the non identity in $C_2$.

I have a few questions.

1) If $\phi:G \rightarrow H$ is a homomorphism what can we say about the order of elements? (I.e. is it true that $o(g)|o(\phi(g))$ or the other way around maybe?)

2) Are my maps correct and are the the only homomorphisms from $V_4$ to $C_2$.

3) Is there a nicer way to argue this question other than just trying to guess all possible ones?

Thanks!

• If $g \in G$ has order $n$, then $\phi(g)^n = \phi(g^n) = \phi(e_G) = e_H$, and so $o(\phi(g))$ divides $n = o(g)$. May 15 '17 at 10:46
• A homomorphism is always fully determined when you know where generating elements go. The delicate questions arise in determining what kind of constraints are there, or can we possibly map the generators any which way we want? $V_4$ is generated by two commuting elements of order two, so the images of those two generators must go to such an element whose square is trivial. This is no constraint, when the target is $C_2$. Nor is commutation. That leaves ____ alternatives. May 15 '17 at 10:51
• And... some of the maps you describe are not homomorphisms. Let $V_4=\{1,a,b,c\}$, $C_2=\{1,g\}$. If $a\mapsto 1$ and $b\mapsto g$, then $c=ab$ must $\mapsto 1g=g$. May 15 '17 at 10:52
• @JyrkiLahtonen So since $V_4=\langle (12)(34), (13)(24)\rangle := \langle a,b \rangle$ then any homomorphism $\Phi$ is determined by where $\Phi$ maps $a$ and $b$. Here we can have $3$ possible choices $a,b$ are both mapped to $1$ giving the trivial homomorphism. Alternatively we have $a$ is mapped to $1$ and $b$ is mapped to $g$ or vice versa. So in total there are $3$ homomorphisms from $V_4$ into $C_2$?Is that correct? May 15 '17 at 12:39
• @RyanS A good start! Nothing stops you from mapping both $a$ and $b$ to $g$, though :-) May 15 '17 at 12:51

The Klein four group $V_4$ is isomorphic to $\mathbb{Z}_2\oplus \mathbb{Z}_2.$ $C_2$ is $\mathbb{Z}_2$. Since the groups are abelian homomorphisms from $\mathbb{Z}_2\oplus \mathbb{Z}_2$ to $\mathbb{Z}_2$ are $\mathbb{Z}$-linear maps $\text{Hom}_{\mathbb{Z}}(\mathbb{Z}_2\oplus \mathbb{Z}_2, \mathbb{Z}_2)$. Since $\text{Hom}_{\mathbb{Z}}(\mathbb{Z}_2,\mathbb{Z}_2) = \mathbb{Z}_2$ we have \begin{eqnarray} \text{Hom}_{\mathbb{Z}}(\mathbb{Z}_2\oplus \mathbb{Z}_2, \mathbb{Z}_2) & \cong & \text{Hom}_{\mathbb{Z}}(\mathbb{Z}_2, \mathbb{Z}_2) \oplus \text{Hom}_{\mathbb{Z}}(\mathbb{Z}_2, \mathbb{Z}_2)\\ & = & \mathbb{Z}_2 \oplus \mathbb{Z}_2. \end{eqnarray}
If you have already done linear algebra, you may notice that $V_4$ is a $C_2$ vector space with dimension $2$. This allows you to write all the functions as matrices of dimension $2×1$, since being an homomorphism is the same as being linear (check it!). Indeed, the only choices you can take is where to send the generators.
Moreover, always given linear algebra topics, you're evaluating the cardinality of the dual space, that is isomorphic to $V_4$.
By the number of ways $C_2$ is partitioned into the quotient subgroups of (together with maybe lagrange theorem) of $C_2$ which are isomorphic to the subgroups of $V_4$. By the isomorphism theorem. So the number of subgroups of $V_4$.
The group isomorphism theorem says for every sub group of $V_4$ there is a homomorphism $V_4\to C_2$ and the subgroup of $V_4$ is isomorphic to the quotient group of $C_2$. Thus one can think of the homorohisms from $V_4\to C_2$ as essentially the subgroups and vice versa. Homomorphisms up to isomorphism that is.