Consider $\rho = (1; 2; 3; 4)$ and $\tau = (2; 3)(1; 4)$ in $S_4.$

Question

(a) Prove that $\rho^4 = (1)$ and $\tau^2 = (1).$

(b) Prove that $\tau \rho = \rho^{−1}\tau.$

For part $a$ I am not sure why I can't get $(1)$ when I start mapping things. Am I missing something? It seems like this would be simple.

Answer 1

Hope you understand the cycle representation of the permutations given. Let us see what $ \rho $ does by applying it to a set of four symbols ${1,2,3,4}$

$\rho (1,2,3,4) = (4,1,2,3)$

$\rho^{2} (1,2,3,4) = \rho ( \rho (1,2,3,4)) =\rho (4,1,2,3) =(3,4,1,2) $

$\rho ^{3}(1,2,3,4) =\rho (3,4,1,2) =(2,3,4,1) $

$\rho^{4}(1,2,3,4) =\rho (2,3,4,1)= (1,2,3,4) $

So $\rho^{4} $ is just the identity map.

Similarly let us see what $\tau $ does

$\tau(1,2,3,4)= (4,3,2,1) $

$\tau^{2}(1,2,3,4)= \tau(\tau(1,2,3,4)=\tau(4,3,2,1)=(1,2,3,4) $

So $\tau^{2} $ is just the identity map

Answer 2

A permutation $\sigma\in\mathfrak{S}_n$ is a mapping $\sigma:\{1,\dots,n\}\to\{1,\dots,n\}$. The permutation $\rho=(1,2,3,4)$ is the mapping associating $i\mapsto i+1\, (\text{mod } 4)$.

What is $\rho^n$? It is the composition of $\rho$ with itself $n-$times. Therefore $\rho^n$ is the mapping associating $i\mapsto i+n\, (\text{mod } 4)$. When $n=4$, or more generally $n=4k$, the mapping $\rho^{4k}$ associates $i$ to $i+4k=i\, (\text{mod } 4)$. Hence $\rho^4$ is nothing else than the identity.

The case of $\tau$ is similar. $\tau$ is the product of two disjoint cycles, hence $\tau^2=(1,2)(3,4)(1,2)(3,4)=(1,2)(1,2)(3,4)(3,4)$ that is the identity by the previous observation. Remember that disjoint cycles commutes!

Answer 3

An $n$-cycle always has order $n$. And it's pretty easy to see why. Since $(abcdefgh)$ means $a\to b\to c\to d\to e\to f\to g\to h\to a$, and in this example we have an $8$ cycle, just to illustrate. Apply twice and you get $a\to c\to e\to g\to a$ and $b\to d\to f\to h\to b$. So each additional time you apply the $n$-cycle, you "skip" one more place. So, when you skip $n$ places, you wind up with the identity.

Now, for the second one, when you have a product of disjoint cycles, the order is the $\operatorname{lcm}$ of the lengths. So the order of a product of two disjoint transpositions is $2$.