Question

(a) If $\mathit{a}$ and $\mathit{b}$ each have order 3 in a group and role="math" localid="1654346029946" ${\mathit{a}}^{\mathbf{2}}\mathbf{=}{\mathit{b}}^{\mathbf{2}}$, prove that role="math" localid="1654346100205" $\mathit{a}\mathbf{=}\mathit{b}$.

Short answer

We proved that,$a=b$.

Step-by-step solution

To find another identity element

Let $G$be a group and $a,b\in G$ be any two elements of group $G$.

Thedefinition of the order of groupstates 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 $\mathit{G}$group is called the order of $\mathit{G}$.It is denoted by $o\left(G\right)$.

It is given that$o\left(a\right)=o\left(b\right)=3$ .

Assume that $a^2=b^2$.

To show that $a=b$.

Let be the identity element of $G$.

To prove  a=b

Now,

$$\begin{gathered} a=ae \\ =a{a}^3\dots (\because o\left(a\right)=3) \\ \begin{array}{c}=a^4\\ ={\left(a^2\right)}^2\\ ={\left(b^2\right)}^2\\ =b^4\\ =b{b}^3\\ \begin{array}{c}=be\dots (\because o\left(b\right)=3)\\ =b\end{array}\end{array} \end{gathered}$$

$\therefore a=b$

Hence, .$a=b$