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I'm trying to understand the proof for

Let $\sigma$ be any element of $S_n$. Then $\sigma$ may be expressed as a product of disjoint cycles.

In this proof (p. 3) there's a part where $p=\sigma \tau^{-1}$ is said to "fix" each element of the set $\{a_i : i≤j\}$. Here $\tau$ is the k-cycle $(a_1, a_2, ..., a_j)$ and $\sigma$ is a function $\sigma(a_i)=a_{i+1}$.

What does it mean for $p$ to "fix" each element? How does this happen?

mavavilj
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    A mapping $f \colon X \to X$ fixes an element $x \in X$ if $f(x) = x$. We also call $x$ a fixed point (or sometimes fixpoint) of $f$. – M. Vinay Jun 03 '16 at 12:11
  • "Fixes the element $a_1$" means "Does not move the element $a_1$", or "Keeps the element $a_1$ fixed". – Arthur Jun 03 '16 at 12:14

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The permutation $\tau$ maps each $a_i$ to $a_{i+1}$, for $1 \le i < j$, and maps $a_j$ to $a_1$. Its inverse $\tau^{-1}$ maps each $a_i$ to $a_{i-1}$, for $1 < i \le j$, and maps $a_1$ to $a_j$.

On the other hand, $\sigma(a_i) = a_{i+1}$, by definition of $a_i$, for all $i \ge 1$. Therefore, $$\sigma(\tau^{-1}(a_i)) = \sigma(a_{i-1}) = a_i,$$ for each $i \le j$ (read $i -1$ as $j$ for $i = 1$). Thus, $\sigma \tau^{-1}$ maps each $a_i$ to $a_i$, or in other words, fixes each $a_i$, for $i \le j$.

M. Vinay
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