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Define the action of $S_n$ on $\mathbb{R}^n$:

take any $x\in S_n$, consider the mapping $x: \mathbb{R}^n\to\mathbb{R}^n$, $e_1, e_2 ...e_n$ are the standard basis of $\mathbb{R}^n$, $x(e_k)=e_{x(k)}$, clearly $x$ is a linear transformation.

How to show that if $x\in S_n$ is a reflection to some vector then $x$ is a transposition?

I can only show that transpositions are reflections, how to show that there are no more?

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2 Answers 2

If $x \in S_n$ defines a reflection of $\mathbb R^n$, then $x$ has order 2, thus it is a product of transpositions. If one of these transpositions is $(1\;2)$ then $e_1$ is mapped to $e_2$ and vice versa. There is only one reflection with this property, namely the reflection at the hyperplane orthogonal to $e_1-e_2$. This reflection leaves all the $e_i$ ($i \neq 1,2$) unchanged, so $x$ is the transposition $(1\;2)$.

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This is a great explanation. OP should accept this :) –  Enzo Apr 24 '13 at 5:19
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If $p$ is a permutation whose disjoint cycle decomposition is $c_1c_2\cdots c_r$ (including cycles of length $1$), then the multiplicity of $1$ as an eigenvalue of the map induced by $p$ is precisely $r$. It follows that $p$ is a reflection iff there is exactly one cycle of length $2$ and all the rest are of length $1$; of course, this happens exactly when $p$ is a transposition.

Of course, we have to check my initial claim on the multiplicity of $1$.

Consider the matrix $M$ of the map induced by $p$. It is easy to see that, up to permuting the basis, $M$ is in fact a diagonal block matrix, with one block corresponding to each of the cycles $c_i$. The multiplicity of $1$ as an eigenvalue is therefore the sum of the multiplicities of $1$ as eigenvalues of each of the permutation matrices corresponding to the $c_i$. We are thus reduced to showing that if $p$ is the cycle $(1,2,3\dots,n)$, then $1$ is a simple eigenvalue. But this is immediate: the characteristic polynomial, which coincides with the minimal one, is $x^n-1$.

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