# How to prove that a given function is not a group homomorphism?

How to disprove a group homomorphism?

$\text{For } n \in \mathbb N, \pi \in S_n \text{is } S_n \rightarrow S_n, \sigma \mapsto \pi \sigma \text{ a group homomorphism}$.

I would like to prove that this is wrong.

$\phi(xx´) = \pi \sigma \pi \sigma ´ \text{ and } \phi(x) \phi(x´)= \pi \sigma \pi \sigma´\text{ so: } \phi(xx´)= \phi(x) \phi(x´)$. Well, I see that I have proved the opposite but I have no idea how I could do it the right way.

• Why do you say that $\phi(xx´) = \pi \sigma \pi \sigma$? Apr 3, 2017 at 7:07
• To disprove something is a homomorphism, you just need to find 2 elements (or one repeated) that the homomorphism property fails. Try a specific $n,\pi$ and elements and see what happens
– Alan
Apr 3, 2017 at 7:08
• @ChrisCulter Should it be $\phi(xx´)= \pi \sigma \sigma ´$? Apr 3, 2017 at 7:16
• @jublikon Yep! Assuming, that is, that you're setting $x=\sigma$ and $x'=\sigma'$. It would be clearer to write just $\phi(xx')=\pi xx'$. Apr 3, 2017 at 7:19
• @jublikon Right! If $\pi$ is anything else, the identity fails, so your argument successfully proves that $\phi$ is not a homomorphism. Apr 3, 2017 at 7:37

A group homomorphism will always take the identity to the identity, but the given function takes the identity to $\pi$.