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I have seen this proof Is any permutation the product of two involutions? but that is unclear to me
This proves it for only the cycle $(1,2,3...n)$
How does proving it only for the cycle $(1, 2, 3...n)$ helps proving for any kind of permutation?

I am new to group theory and I do not know anything other than the following

  1. Linear algaebra
  2. Any permutation can be written as a product of transpositions.
  3. Any permutation can be written as a product of disjoint cycles.
  4. A permutation is an involution precisely if it can be written as a product of one or more non-overlapping transpositions.
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    $\begingroup$ By 3, Any permutation can be written as the product of disjoint cycles. So, if you can prove that a cycle is the product of 2 involutions, then you can show that the permutation is the product of disjoint "product of 2 involutions", which hopefully you can then show is simply the product of 2 involutions. This turns out to be true because the product of several involutions which have disjoint terms is still an involution. Putting it all together, $ P = \prod C_i = \prod A_i B_i = (\prod A_i) ( \prod B_i) $ gives the permutation as the product of 2 involutions. $\endgroup$ – Calvin Lin Jun 26 '20 at 7:24
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It might help to consider an example. Suppose we have the following disjoint cycle decomposition of a permutation: $$ \sigma = (1 \ 7\ 4\ 10)(2\ 9\ 8)(3\ 5\ 6). $$ To begin, separately decompose each cycle: $$ (1 \ 7\ 4\ 10) = \tau_{1,1}\tau_{1,2}, \quad (2\ 9\ 8) = \tau_{2,1}\tau_{2,2} \quad (3\ 5\ 6) = \tau_{3,1} \tau_{3,2}. $$ Note that $\tau_{1,1},\tau_{1,2}$ are permutations that only affect the elements $1,4,7,10$. Similarly, $\tau_{2,1},\tau_{2,2}$ only affect $2,9,8$. In other words, for any $i \neq j$, the elements $\tau_{i,p}, \tau_{j,q}$ are disjoint permutations, which means that $\tau_{ip}\tau_{jq} = \tau_{jq}\tau_{ip}$.

With that established, we can use this commutativity property to "move" the transpositions $\tau_{i,1}$ to the left. That is, we can write $$ \sigma = \tau_{11}(\tau_{12}\color{red}{\tau_{21}})\tau_{22}\tau_{31}\tau_{32}\\ = \tau_{11} (\color{red}{\tau_{21}}\tau_{12}) \tau_{22}\color{red}{\tau_{31}} \tau_{32}\\ = \tau_{11}\tau_{21}\color{red}{\tau_{31}}\tau_{12}\tau_{22}\tau_{32}\\ = (\tau_{11}\tau_{21}\tau_{31})(\tau_{12}\tau_{22}\tau_{32}). $$ Now, we see that $\sigma$ is a product of the involutions $\tau_{11}\tau_{21}\tau_{31}$ and $\tau_{12}\tau_{22}\tau_{32}$.

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