# $\exists \phi:S_n \to Aut(S_n)$ homomorphism and $1-1$,$n\geq 3$

Question

Show that $\exists \phi:S_n \to Aut(S_n)$ homomorphism and $1-1$,when $n\geq 3$

Attempt

Let $\phi(\sigma)=f_{\sigma}$ with $f_{\sigma}:S_n\to S_n$ with $f_{\sigma}(t)=\sigma t\sigma^{-1}$.Now:

$1)\phi$ is well defined:we have to show that $f_{\sigma}\in Aut(S_n)$ which I think I proved correctly

$2)\phi$ is homomorphism:$\phi(\sigma_1\sigma_2)=f_{\sigma_1\sigma_2}$.But $f_{\sigma_1\sigma_2}(t)=(\sigma_1\sigma_2)t(\sigma_1\sigma_2)^{-1}=f_{\sigma_1}\circ f_{\sigma_2}(t)$,so $f_{\sigma_1\sigma_2}=f_{\sigma_1}\circ f_{\sigma_2}=\phi(\sigma_1)\phi(\sigma_2)$

$3)\phi$ is $1-1$:$\phi(\sigma_1)=\phi(\sigma_2)\Rightarrow \sigma_1 t\sigma_1^{-1}=\sigma_2t\sigma_2^{-1} \forall t\in S_{n}\Rightarrow \sigma_2^{-1}\sigma_1\in Z(S_n)=\{1_{S_{n}}\}$ because $n\geq3$ so $\sigma_1=\sigma_2$

Is this correct?Is there a quicker way to prove this?

• It looks fine to me. – Javi Feb 27 '18 at 17:22
• thank you for checking it! – giannispapav Feb 27 '18 at 17:25