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.

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    $\begingroup$ Why do you say that $\phi(xx´) = \pi \sigma \pi \sigma$? $\endgroup$ Apr 3, 2017 at 7:07
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    $\begingroup$ 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 $\endgroup$
    – Alan
    Apr 3, 2017 at 7:08
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    $\begingroup$ @ChrisCulter Should it be $\phi(xx´)= \pi \sigma \sigma ´$? $\endgroup$
    – jublikon
    Apr 3, 2017 at 7:16
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    $\begingroup$ @jublikon Yep! Assuming, that is, that you're setting $x=\sigma$ and $x'=\sigma'$. It would be clearer to write just $\phi(xx')=\pi xx'$. $\endgroup$ Apr 3, 2017 at 7:19
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    $\begingroup$ @jublikon Right! If $\pi$ is anything else, the identity fails, so your argument successfully proves that $\phi$ is not a homomorphism. $\endgroup$ Apr 3, 2017 at 7:37

1 Answer 1


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

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    $\begingroup$ Note that when the permutation is the identity, the function is indeed a homomorphism. $\endgroup$ Apr 3, 2017 at 19:42

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