Show that there are exactly two homomorphisms $\alpha:C_6\rightarrow C_4$.

Question

Show that there are exactly two homomorphisms $\alpha:C_6\rightarrow C_4$.

My first thought is to find the two homomorphisms, thus proving that there are at least two homomorphisms. Then I would need to prove that there are at most two.

I believe I have found one, that is $\alpha:x\mapsto x^3$. It works for a few elements in $C_6$. Though before I do all 36 possibilities and look for another homomorphism, is there a more efficient way to show that there are at least two? Also I am not sure where to begin showing that there is at most two.

Answer 1

Let $g$ be a generator of $C_6$. Then $\alpha$ is completely determined by $h=\alpha(g)$.

Now $1=\alpha(1)=\alpha(g^6)=\alpha(g)^6=h^6$. Also, $h^4=1$.

So $h^d=1$, for $d=\gcd(4,6)=2$.

The question thus reduces to finding all elements of order at most $2$ in $C_4$. Each such element defines a homomorphism $C_6\to C_4$.

Answer 2

Let $C_6 = \{1,x,\ldots,x^5\}$ and let $C_4 = \{1,y,\ldots, y^3\}$. If $\alpha$ is a homomorphism, then $\alpha(x^k)=\alpha(x)^k$, so determining the value of $\alpha(x)$ completely determines how $\alpha$ acts on the other elements of $C_6$.

The other restriction we need to consider is that $\alpha(x)$ must satisfy $1=\alpha(1)=\alpha(x^6)=\alpha(x)^6$. Which elements of $C_4$ satisfy this?

Answer 3

Remember that there is a 1-1 correspondence between the normal subgroups of a group and the kernels of group homomorphisms. In other words, every normal subgroup of a group can be realized as the kernel of some homomorphism.

So for your problem, first examine the normal subgroups of $C_6$ (which is all of them since $C_6$ is abelian). This should give you some insight about the number of homomorphisms between $C_6$ and $C_4$.