# automorphisms of symmetric group.

Prove that for each $\displaystyle \sigma \in Aut(S_n)$
$\displaystyle \sigma: (1 \, 2) \mapsto (a \, b_2), \, \sigma: (1 \, 3) \mapsto (a \, b_3), \ldots, \sigma: (1,\, n) \mapsto (a \, b_n)$.
I know that an automorphism sends a transposition to a transposition so $\displaystyle \sigma: (1 \, 2) \mapsto (a \, b_2)$ is fine. How to prove the rest. I tried to use contradiction. denote $\displaystyle \sigma((1 \, 3))= \tau$. Assume $\displaystyle \tau (a)=a, \tau (b_2)=b_2$. Then we should get a contradiction but i couldn't arrive at one.
i just found something. let $\displaystyle \sigma((1 \, 2))=(a \, b), \sigma((1 \, 3))= (c \, d)$. Consider $\displaystyle \sigma((1 \, 2)(1 \, 3))=\alpha$. So $\displaystyle \sigma((1 \, 3 \, 2))= \alpha \Rightarrow |\alpha|=3$ since $\displaystyle \sigma$ is an isomorphism. Also $\displaystyle \sigma((1 \, 2)(1 \, 3))=\sigma((1 \, 2)) \sigma((1 \, 3))=(a \, b)(c \, d)=\alpha$. if $\displaystyle (a \, b) \, & (c \, d)$ are disjoint cycles then $\displaystyle |\alpha|=2$ which is not possible. Hence there is one and only one element common in the cycles $\displaystyle (a \, b) \, & \, (c \, d)$ so WLOG $\displaystyle (c \, d)=(a \, d)$. I guess this will prove the proposition. Please comment.