# Every permutation can be written as a product of transpositions

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?

• 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. – Arturo Magidin Aug 12 at 3:05
• 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. – Derek Elkins Aug 12 at 3:08
• But why did they mention that every point is moved except for b? – Tim Aug 12 at 3:08
• 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 $1$". – Derek Elkins Aug 12 at 3:13
• 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. – Brian Moehring Aug 12 at 3:16