Question

Show that$A_4$ (which has order 12 by Theorem 7.29) has exactly three elements of order 2.

Short answer

It is shown that, $A_4$has exactly 3 elements of order 2.

Step-by-step solution

Obtain that A4have three elements of order 2

Consider the element $A_4$as follows:

$$\begin{gathered} \left(1\right) \\ \left(12\right)\left(34\right)\mathbf{,}\left(13\right)\left(24\right)\mathbf{,}\left(14\right)\left(23\right) \\ \left(123\right)\mathbf{,}\left(132\right)\mathbf{,}\left(124\right)\mathbf{,}\left(142\right)\mathbf{,}\left(134\right)\mathbf{,}\left(143\right)\mathbf{,}\left(234\right)\mathbf{,}\left(243\right) \end{gathered}$$

Result

The order of a 2-cycle in the symmetric group has order 2.

As order of the product two 2-cycle is the greatest divisor of their order, and the order of the product two 2-cycle is 2 as well.

Hence, it follows the list from above the number of element $A_4$of order 2 is 3.