Question

If $\mathbf{\sigma }$ is a k-cycle with k odd, prove that there is a cycle$\mathit{\tau }$ such that ${\mathit{\tau }}^{\mathbf{2}}\mathbf{=}\mathbf{\sigma }$.

Short answer

It is proved that there is a cycle$\tau$ such as that ${\mathrm{\tau }}^2=\mathrm{\sigma }$.

Step-by-step solution

Statement of Lemma 7.27

Lemma 7.27states that the identity permutation in ${\mathrm{S}}_{\mathrm{n}}$would be even, however, not odd.

Show that there is a cycle τ such that τ2=σ

It is observed that for a generic $k$cycle $k=(b_1\cdots b_k)$for role="math" localid="1654329340680" $k$odd, there is $k^2=(b_1{b}_3\dots b_k{b}_2{b}_4\dots b_{k-1})$.

Consider that $\sigma =(a_1{a}_2\dots a_k)$for $k$odd, and let role="math" localid="1654329474539" $\tau :=\left(a_1{a}_{\frac{k+3}{2}}{a}_2{a}_{\frac{k+5}{2}}\cdots a_{k-1}{a}_{\frac{k-1}{2}}{a}_k{a}_{\frac{k+1}{2}}\right)$.

In other words, construct role="math" localid="1654329508431" $\tau$be a concatenation of role="math" localid="1654329552366" $a_i{a}_{\frac{k+2i+1}{2}}$with $1\le i\le \frac{k-1}{2}$and then end the cycle by $a_{\frac{k+1}{2}}$. Then, there is ${\tau }^2=\sigma$.

Hence, it is proved that there is a cycle$\tau$ such as that ${\tau }^2=\sigma$.