I'm confused with how a textbook presents their proof on how every permutation in a permutation group can be represented as a product of transpositions. They said the following:

"Let $\alpha \in S_n$ be a permutation and let $m$ be the number of points moved by $\alpha$. Suppose that $m = 0$. Then $\alpha = ()$. Now suppose that $m \gt 0$. Let $a$ be a single point moved by $\alpha$ and suppose that $b = a^{\alpha}$. Let $\tau = (a, b)$. Then the number of points moved by $\tau^{-1} \alpha$ is the number of points moved by $\alpha$, except for just $b$."

The only problem in this I can see is that what if $\alpha$ itself is equal to $(a,b)$? If $\alpha = (a,b)$, then $\alpha$ moves $a$ to $b$. But then $\tau^{-1} \alpha$ doesn't move $a$ at all. So what is going on here?

  • 1
    $\begingroup$ Nothing; by induction, you get that $\tau^{-1}\alpha$ can be written as a(n empty) product of transpositions, and hence that $\alpha=\tau$, so $\alpha$ is a product of transpositions. $\endgroup$ – Arturo Magidin Aug 12 at 3:05
  • $\begingroup$ The whole point of $\tau^{-1}\alpha$ is to produce a permutation that doesn't move $a$. That's why the number of points moved is less and thus we can apply the induction hypothesis. $\endgroup$ – Derek Elkins Aug 12 at 3:08
  • $\begingroup$ But why did they mention that every point is moved except for b? $\endgroup$ – Tim Aug 12 at 3:08
  • $\begingroup$ The "except for just $b$" may just be a typo that should be "except for just $a$", though that's a bit weird too since we're talking about a count of points moved, not specific points. They should have just said "minus at least $1$". $\endgroup$ – Derek Elkins Aug 12 at 3:13
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    $\begingroup$ Your criticism is valid, but ultimately it doesn't matter. If you want, you can replace that part with the phrase "Then the number of points moved by $\tau^{-1}\alpha$ is at most one less than the number of points moved by $\alpha$" ................ And for those confused about which point is fixed, it looks like this author might define the permutations to act on the right, in which case $b$ is fixed. If you consider the permutations to act on the left, $a$ would be fixed. Either way, the induction still works. $\endgroup$ – Brian Moehring Aug 12 at 3:16

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