Question

Show that ${\mathit{A}}_{\mathbf{3}}$ is a cyclic group of order 3 and hence simple by Theorem 8.25.

Short answer

It is proved that,$A_3$ is a cyclic group of order 3 and$A_3$ is simple.

Step-by-step solution

Determine that A3

Consider $A_3$.

The order of $A_n$is$\frac{n!}{2}$ .

Applying formula

Now, applying this to the order of $A_3$as:

$\begin{array}{c}\frac{3!}{2}=\frac{3\times 2}{2}\\ =\frac{6}{2}\\ =3\end{array}$

Thus,$A_3=3$ .

Therefore,$A_3$ is a cyclic group of order 3 as the group of order 3 is${\mathrm{\mathbb{Z}}}_3$ .

Hence, by theorem 8.25$A_3$ is simple.

Thus, it is proved that$A_3$ is a cyclic group of order 3 and$A_3$ is simple.