How many group homomorphisms $\phi: S_5\to C_5$ are there?

Denote by $S_5$ and $C_5$ the fifth symmetric and cyclic group respectively. How many homomorphisms $\phi: S_5\to C_5$ are there? I think that there is only one, but I'm not too confident in my attempted proof!

My attempted proof: for $i= 1,...,4$, denote the transposition $\sigma_{i,i+1}:=(i\;\; i+1)$. Since $\sigma_i$ generate $S_5$, all homomorphisms, $\phi$, are determined by their behaviour on $\sigma_i$. Observe that $\sigma_i^2 = \mathrm{Id}\implies \phi(\sigma_i)^2=\mathrm{Id}$ and that $\phi(\sigma_i)^5 = \mathrm{Id}$, since the target group is $C_5$. Therefore, the order of $\phi(\sigma_i)$ must be 1 for all $i$. Hence there is only one homomorphism, namely $\phi\equiv \mathrm{Id}$.

Solution 1:

Whenever looking at a homomorphism from a larger group to a smaller group, it is wise to also ask "what do I know about the subgroup structure of my domain?"

Any map $\phi:S_5\longrightarrow C_5,$ will have to have $A_5,$ the commutator subgroup of $S_5,$ in the kernel. Since $S_5/A_5\cong \Bbb{Z}/2\Bbb{Z},$ then one asks if $C_5$ has any elements of order 2, and the answer is certainly no, since $C_5$ is generated by any of its non-identity elements, and thus any such homomorphism must be as you described in your answer.

Solution 2:

Your solution is fine. Here is another approach.

$\ker \phi$ is a normal subgroup of $S_5$, which has only three normal subgroups: $1$, $A_5$, $S_5$.

• $\ker \phi \ne 1$ because $S_5$ has $120$ elements and $C_5$ has only $5$.

• $\ker \phi \ne A_5$ because $A_5$ has index $2$ but $C_5$ has no subgroup of order $2$.

Therefore, the only possibility is $\ker \phi = S_5$. In other words, $\phi$ is the trivial homomorphism.