Inner automorphisms and the permutation group $S_n$

Question

Prove that $Inn(S_n) \cong S_n$, where $Inn(S_n)=\{a_s|s \in S_n\}$ and $a_s(t)=s^{-1}ts$ for $t \in S_n$

Proof:

We define a function $\Psi :S_n \longrightarrow Inn(S_n)$ such that $\Psi(s)=a_s$

1)If $s_1=s_2$ then $a_{s_1}(t)=s_1^{-1}ts_1=s_2^{-1}ts_2=a_{s_2}(t)$ ,$\forall t \in S_n$, thus $\Psi(s_1)=\Psi(s_2)$ proving that $\Psi$ is well defined.

2)$\Psi$ is a homomorphism because $\Psi(s_1s_2)=a_{s_1s_2}$ and $\forall t \in S_n$ $a_{s_1s_2}(t)=s_2^{-1}s_1^{-1}ts_1s_2=s_2^{-1}a_{s_1}s_2=a_{s_2}(a_{s_1})(t)$ so we have that $\Psi(s_1s_2)=\Psi(s_2) \Psi(s_1)$

3)$\Psi$ a monomorphism.

$\Psi(s)=a_{id}$ if and only if $\forall t \in S_n,a_s(t)=t$ if and only if $\forall t \in S_n$,$s^{-1}ts=t$ if and only if $\forall t \in S_n ,ts=st$ if and only if $s \in Z(S_n)={id}$ thus $Ker \Psi=\{id\}$

4)If $b \in Inn(S_n)$ then $b=a_s$ for some $s \in S_n$ and $b=a_s=\Psi(s)$ proving that $\Psi$ is an epimorphism.

Is this proof correct or am i missing something?

Answer 1

This almost works, but your $\Psi$ is actually not a homomorphism! You found that $\Psi(s_1s_2)=\Psi(s_2)\Psi(s_1)$, but if it were a homomorphism it should satisfy $\Psi(s_1s_2)=\Psi(s_1)\Psi(s_2)$ instead. See if you can tweak your definition of $\Psi$ so that $\Psi(s_1s_2)$ will be $\Psi(s_1)\Psi(s_2)$ rather than $\Psi(s_2)\Psi(s_1)$.

Also, in proving $\Psi$ is injective, you claimed that $Z(S_n)$ is trivial. This is only true if $n>2$. In fact, the result you are trying to prove is only true if $n\neq 2$.