Please refer to the question posted here $\operatorname{Aut}(S_4)$ is isomorphic to $S_4$
Here, I am not able to prove how $f$ fixes $3$-cycles.
My attempt –
We have to prove that $\ker(\phi)=\operatorname{id}_{S_4}$. Let $\phi(f)=(1)\in S_4$ which implies $f(P_i)=P_i \ \forall i=1,2,3,4. $ We have to prove $f\operatorname{id}_{S_4}$.
Let $(a\ b\ c )$ be a $3$– cycle. Then $\langle(a\ b\ c )\rangle$ is one of $P_i$'s where $\langle(a\ b\ c )\rangle$ is the subgroup generated by $(a\ b\ c )$. So, $\phi(f)(\langle(a\ b\ c )\rangle)=\langle(a\ b\ c )\rangle$ and therefore $f\bigl(\langle(a\ b\ c )\rangle\bigr)=\langle(a\ b\ c )\rangle$. This implies that $f((a\ b\ c ))=(a\ b\ c )$ or $f \bigl((a\ b\ c) \bigr) =(a\ b\ c )^2=(a\ c\ b )$. Here I am not able to eliminate the case $f\bigl((a\ b\ c )\bigr)=(a\ c\ b )$.
Can anyone please help me?
Thanks!
Best Answer
Assume $f$ acts as identity on $P_i$ and as inversion on $P_j$. Let $d$ be the point fixed by $P_i$ and $c$ the point fixed by $P_j$, and $a,b$ the other two. Then $$f((a\,c)(b\,d))=f(\underbrace{(a\,b\,c)}_{\in P_i}\underbrace{(a\,b\,d)}_{\in P_j})= (a\,b\,c)(a\,d\,b)= (a\,d\,c),$$ which is impossible (order 2 $\ne$ order 3).
And if $f$ acts as inversion on both $P_i$ and $P_j$, $$ f((c\,b\,d))=f((a\,c\,b)(a\,b\,d))=(a\,b\,c)(a\,d\,b)=(a\,d\,c),$$ so $f$ maps $\langle (b\,c\,d)\rangle$ to $\langle (a\,c\,d)\rangle$, contradiction.
We conclude that the only possibility is that $f$ acts as identity on all $P_i$.