# 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 $\phi\equiv \mathrm{Id}$.

• I think it is fine. May 7 '17 at 16:48
• Yes, your proof is a very nice way to do this. May 7 '17 at 16:51
• This question deserves upvotes not only because is right but because the OP took the time to actually show his own effort, thoughts and doubts. And the written proof is very simple, elementary...and, of course, correct. +1 May 7 '17 at 16:57
• My hope for the poster of this question, is that his confidence is hereby boosted. May 7 '17 at 17:03
• Keep up the good work! May 7 '17 at 17:36

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.

• I actually like the proof given by the OP a lot more, as it takes much less theory. May 7 '17 at 16:51
• I'm not posting this because the OP is wrong in any way. I was merely attempting to offer some insight which will help when computing things directly isn't so clear. May 7 '17 at 16:52
• Although the OP may or may not realize it, his proof does something similar to (but with the co-domain) what you suggest: it uses the fact that $C_5$ is simple (since $5$ is prime), to show that all subgroups of $S_5$ of order two, map to the identity of $C_5$ (since $2\not\mid 5$). In particular, $S = \{\sigma_i:i = 1,2,3,4\}$ maps to $e_{C_5}$, so that $\phi(S_5) = \phi(\langle S\rangle) = \langle \phi(S)\rangle = \langle e_{C_5}\rangle = \{e_{C_5}\}$. May 7 '17 at 16:58

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.