Question

List all the elements of ${\mathit{A}}_{\mathbf{4}}$ and express each as a product of 3-cycles.

Short answer

Every element of$A_4$ is written as the product of 3-cycles.

Step-by-step solution

Determine the list of all the elements of A4

Consider the elements of $A_4$.

Now, consider the fact and prove it.

For $n\ge 3$, every element of $A_4$is written as the product of 3-cycles.

At the starting point, every element in$A_4$ is written as a product of even number of transposition, which is enough to show that for$n\ge 3$ and any transpositions$\mathit{\sigma }\mathbf{,}\mathit{\tau }$ in$A_n$ ,$\sigma \tau$

is written as product of 3-cycles.

Determine Case I and Case II

Case I ( $\sigma$and $\tau$are disjoint)

Let’s consider that $\sigma =\left(ij\right)$and $\tau ==\left(kl\right)$ for $i,j,k,l\in \{1,\mathrm{....},n\}$ with $i\ne j\ne k\ne l$.

Then, $\sigma \tau =\left(ij\right)\left(kl\right)$ implies $\sigma \tau =\left(ijk\right)\left(jkl\right)$.

It is verifying that, $\sigma \tau$ is written as a product of 3-cylces when $\sigma$and $\tau$are disjoint transposition.

Case II ( $\sigma$and $\tau$are not disjoint)

Let’s consider that $\sigma =\left(ij\right)$and $\tau =\left(jk\right)$ for $i,j,k,l\in \{1,\mathrm{....},n\}$ with $i\ne j\ne k\ne l$.

Then, $\sigma \tau =\left(ij\right)\left(jk\right)$ implies $\sigma \tau =\left(ijk\right)$.

It is verifying that,$\sigma \tau$ is written as product of 3-cylces when$\sigma$ and$\tau$ are not disjoint transpositions.

Thus, the above claim is verified.

Result

Now, apply this fact with $n=4>3$.

From this, it is concluded that, every element of$A_4$ is the written as a product of 3-cycles.