Question

(b) If $\mathit{G}$is a finite group, prove that there is an even number of elements of order 3 in $\mathit{G}$ .

Short answer

We proved that the number of elements of order 3 is$2(n+1)$ in $G$.

Step-by-step solution

To prove the result for two elements

Let $G$be a group and $x_0\in G$be an element of group $G$.

The definition of the order of group states that:

Let $\mathit{G}$be a group. Then $\mathit{G}$ is said to be a finite group (or of finite order) if it has a finite number of elements. It is also called the cardinality of the group.

Here, the number of elements in group $\mathit{G}$ is called the order of $\mathit{G}$.It is denoted by $o\left(G\right)$.

Then $o\left(x_0\right)=3$.

Now, ${\left(x_0^2\right)}^3={\left(x_0^3\right)}^2=e$

$\Rightarrow o\left(a_0^2\right)=3$

Let $a_1\in G\backslash \left\{a_0\right\}$ such that $o\left(a_1\right)=3$.

From part (a), we can say that $a_0^2\ne a_1^2$ .

$\therefore \{a_0,a_0^2\}\cap \{a_1,a_1^2\}=\varphi$

To prove there is an even number of elements of order 3

In this way, we can conclude that elements $\{a{}_i,0\le i\le n\}$such that $o\left(a_i\right)=3,\forall i$, where $a_i\in G\backslash \{a_0,a_1,a_2,\dots ,a_{n-1}\}$.

Thus, $\underset{i=0}{\overset{n}{\cup }}\{a_i,a_i^2\}$, which is the set elements of order 3.

$\therefore \{a_i,a_i^2\}\cap \{a_j,a_j^2\}=\varphi$, whenever $i\ne j$.

Hence, the number of elements of order 3 is role="math" localid="1654347820226" $2(n+1)$ in $G$.