Question

Let τ be a fixed odd permutation in . S₁₀ Show that every odd permutation in S₁₀ is

a product of τ and some permutation in A₁₀

Answer 1

Solution:

Proof. Let $\sigma^{\prime}$ be an odd permutation in $S_{n}$ . We must show that there exists an even permutation $\mu \in A_{n}$ such that $\sigma^{\prime}=\sigma \mu$ . Indeed, we may take $\mu=\sigma^{-1} \sigma^{\prime}$ , since as the product of two odd permutations, it is an even permutation, and

$\sigma^{\prime}=\sigma\left(\sigma^{-1} \sigma^{\prime}\right)$

For completeness, let's prove directly that $\sigma^{-1} \sigma^{\prime}$ is even. From the definition of an odd permutation, there exist a finite number of transpositions $\tau_{1}, \ldots, \tau_{m}$ for some odd $m \in \mathbb{N}$ such that

$\sigma=\tau_{1} \ldots \tau_{m}$

Similarly, since $\sigma^{\prime}$ is also an odd permutation, there exist a finite number of transpositions $\tau_{1}^{\prime}, \ldots, \tau_{\ell}^{\prime}$ for some odd $\ell \in \mathbb{N}$ such that $\sigma^{\prime}=\tau_{1}^{\prime} \ldots \tau_{\ell}^{\prime}$ . Consider now the permutation

$\mu=\sigma^{-1} \sigma^{\prime}$

I claim that this lies in $A_{n}$ . Indeed we have

$\mu=\sigma^{-1} \sigma^{\prime}=\underbrace{\tau_{m} \ldots \tau_{1} \tau_{1}^{\prime} \ldots \tau_{\ell}^{\prime}}_{m+\ell}$

The sum of two odd numbers is even, and so it follows that this is an even permutation.

Thus, if we take $n=10$ , we can prove it same way.