Automorphisms of the Symmetric Group

Question

Let $n > 6$. Show that an automorphism $f: S_{n} \to S_{n}$ (an isomorphism from $S_{n}$ to itself) will map transpositions to transpositions.

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I'm not sure on how to start with the problem. I was thinking of using the concept of centralizers (in the case of a transposition $(ij)$, the centralizer will be such $\sigma \in S_{n}$ with $\sigma(i), \sigma(j) \in \{i,j\}$) in order to tackle the problem but thus far I haven't been successful. What would be the correct approach?

Thanks for the help.

Answer 1

It is almost the idea.You know that $f((1,2))$ will be an element of order $2$. Whence a product of $k$ disjoint transposition.

Instead of computing the centralizer of $(1,2)$ and $f((1,2))$ try to compute the number of elements in the conjugacy class (it is much easier). Justify that provided that $n\neq 6$ the only possibility for the numbers of conjugates of $(1,2)$ and of conjugates of $f((1,2))$ is to have $k=1$. To be more precise :

Let $k\geq 1$, denote $T_k$ to be the number of elements conjugated to $(1,2)...(2k-1,2k)$. Show that :

$$T_k=\frac{1}{k!}\begin{pmatrix}n\\2\end{pmatrix}...\begin{pmatrix}n-2k+2\\2\end{pmatrix} $$

Once you have done this, you just need to show that $T_1=T_k$ has for unique solution $k=1$ (provided that $n\neq 6$).