# Permutation equation [closed]

It is given a permutation $$\alpha=\left( \begin{array}{cccccccccc} 1&2&3&4&5&6&7&8&9&10 \\ 7&6&5&9&10&2&1&4&8&3 \\ \end{array} \right).$$ Can equations $$\pi^{29}=\alpha,$$ $$\pi^{30}=\alpha,$$ $$\pi^{31}=\alpha$$ and $$\pi^{32}=\alpha$$ be solved? Can someone explain me the way to solve this task?

• Needs some definitions. What is $\pi$? – motherboard Jan 12 at 5:42
• $\pi$ is also a permutation we have to find. – math_ Jan 12 at 8:14

Here is a starter. It is convenient to look at the cycle notation of $$\alpha$$: \begin{align*} \color{blue}{\alpha=(1\ 7)(2\ 6)(3\ 5\ 10)(4\ 9\ 8)}\tag{1} \end{align*} We see $$\alpha$$ consists of two involutions $$(1\ 7)$$ and $$(2\ 6)$$ which have order two and two cycles of length $$3$$ which have order $$3$$, namely \begin{align*} (1\ 7)^2=\varepsilon\qquad\qquad&(3\ 5\ 10)^2=(3\ 10\ 5)\\ &(3\ 5\ 10)^3=\varepsilon\tag{2}\\ (2\ 6)^2=\varepsilon\qquad\qquad&(4\ 9\ 8)^2=(4\ 8\ 9)\\ &(4\ 9\ 8)^3=\varepsilon\\ \end{align*} with $$\varepsilon$$ the identity permutation.

From (1) and (2) we see $$\alpha^{30}=\varepsilon$$ and consequently \begin{align*} \color{blue}{\alpha^{31}}=\alpha^{30}\circ \alpha=\varepsilon\circ \alpha\color{blue}{=\alpha} \end{align*}

We also derive from (2) \begin{align*} (3\ 10\ 5)^{29}&=(3\ 5\ 10)^{58}=(3\ 5\ 10)^{3\cdot19+1}=(3\ 5\ 10)\\ (4\ 8\ 9)^{29}&=(4\ 9\ 8)^{58}=(4\ 9\ 8)^{3\cdot19+1}=(4\ 9\ 8)\\ (1\ 7)^{29}&=(1\ 7)^{2\cdot 14+1}=(1\ 7)\\ (2\ 6)^{29}&=(2\ 6)^{2\cdot 14+1}=(2\ 6)\\ \end{align*}

It follows: \begin{align*} \color{blue}{((1\ 7)(2\ 6)(3\ 10\ 5)(4\ 8\ 9))^{29}=\alpha} \end{align*}

Permutation can be represented as the product of cycles, where two cycles do not have common elements, as in

$$\alpha=(1\ 7)(2\ 6)(3\ 5\ 10)(4\ 9\ 8)$$

Some properties of permutations and cycles:

The length of a cycle $$\gamma$$ we denote $$l(\gamma)$$. So $$l((1\ 7))=2$$ and $$l((3\ 5\ 10))=3$$. If a permutation is represented by $$n$$ (pairwise disjoint) cycles $$\gamma_1\gamma_2\ldots \gamma_n$$ then $$(\gamma_1\gamma_2\ldots \gamma_n)^k=\gamma_1^k\gamma_2^k\ldots \gamma_n^k\tag {1.1}$$

The permutation $$\gamma_i^k$$ must not be a cycle. It may be the product of more than one cycle. But the following holds, if $$\gamma$$ is a cycle and $$k$$ is a prime:

• If not $$k|l(\gamma)$$, then $$\gamma^k=\delta \tag {1.2}$$ where $$\delta$$ is a cycle and $$l(\delta)=l(\gamma)$$.

• If $$k|l(\gamma)$$, then $$\gamma^k=\gamma_1\gamma_2\ldots \gamma_k \tag {1.3}$$ where all $$\gamma_i$$ are cycles and $$l(\gamma_i)=l(\gamma)/k$$.

If $$\pi$$ is a permutation and $$a,b,c\in \mathbb{Z}$$ then following properties hold:

$$\pi^{ab+c}={(\pi^a)}^b \pi^c\\ \pi^0=\textrm{id}\\ \pi^{-1}=\pi^* \tag {1.4}$$ where $$\textrm{id}$$ is the identity and $$\pi^*$$ is the inverse permutation. If $$\pi$$ is a permutation of $$n$$ elements then $$\pi^{n!}=\textrm{id}$$ holds and if $$\gcd(k,n!)=1$$ then there exists a $$k'$$ such that $$k\cdot k'\equiv 1\pmod{n!} \tag{1.5}$$ and therefore $$(\pi^k)^{k'}=\pi$$.

For a cycle $$\gamma$$ we have $$\gamma^{l(\gamma)}=\textrm{id}$$

Now let's solve $$\pi^{29}=\alpha \tag{2.1}$$

We have $$\alpha^6=\textrm{id}$$ and so $$(\pi^{29})^6=\alpha^6=\textrm{id}$$ and by raising to the power $$29'$$. The symbol $$'$$ is defined in $$(1.5).$$ But $$(\pi^{29})^{29'}=\pi$$ so we get $$\pi^6=\textrm{id}$$ we have$$\pi^{29}=(\pi^6)^{24}\pi^5=\pi^5$$ and so from from $$(2.1)$$ we get $$\pi^5=\alpha$$ $$\textrm{id}=\alpha\pi$$ and $$\pi=\alpha^{-1}$$ So $$\pi=(1\ 7)(2\ 6)(3\ 10\ 5)(4\ 8\ 9)$$ is the unique solution of $$(2.1).$$

The equation $$\pi^{31}=\alpha \tag{2.3}$$ can be processed in a similar way.

To solve $$\pi^{30}=\alpha \tag{2.2}$$

we transform it to $$(\pi^{15})^2=\alpha$$

We first try to solve $$\rho^2=\alpha$$ $$\rho$$ is a square root of $$\alpha$$.

The cycles $$(3\ 5\ 10)(4\ 9\ 8)$$ of $$\alpha$$ can be generated by two cycles of length $$3$$ according to $$(1.2)$$ or by a cycle of length $$6$$ according to $$(1.3)$$ The cycles $$(1\ 7)(2\ 6)$$ of $$\alpha$$ can only be generated by a cycle of length $$4$$

So $$\rho=\phi\psi$$ where either $$\phi=(1\ 2\ 7\ 6)$$ or $$\phi=(1\ 6\ 7\ 2)$$ and where $$\psi=(3\ 10\ 5)(4\ 8\ 9)$$ or $$\psi=(3\ 4\ 5\ 9\ 10\ 8)$$ or $$\psi=(3\ 9\ 5\ 8\ 10\ 4)$$ or $$\psi=(3\ 8\ 5\ 4\ 10\ 9)$$

By combining all possible $$\phi$$ and $$\psi$$ we get $$8$$ different square roots $$\rho$$.

In the next step we want to solve $$\sigma^3=\rho$$ for the square root $$\rho$$.

But $$(1\ 2\ 7\ 6)(3\ 10\ 5)(4\ 8\ 9)$$ does not have a third root because the cycles $$(3\ 10\ 5)(4\ 8\ 9)$$ do not have one according to $$1.1$$ and $$1.2$$ and $$1.3.$$ Similar holds for the other seven square roots $$\rho.$$

In a similar manner it can be shown that $$(2.4)$$ has no solutions.