Question

If ${\mathit{c}}^{\mathbf{2}}\mathbf{=}\mathit{c}$in a group, prove that, c = e.

Short answer

It is proved that,$c=e$

Step-by-step solution

Inverse of an element

Let $\mathit{G}$be a group. An element ${\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathbf{\in }\mathit{G}$is said to be the inverse of an element $\mathit{a}\mathbf{\in }\mathit{G}$if it satisfies $\mathit{a}\mathbf{*}{\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathbf{=}{\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathbf{*}\mathit{a}\mathbf{=}\mathit{e}$, where $\mathit{e}\mathbf{\in }\mathit{G}$is the identity element.

Proof

Given that, $c^2=e$. Now, multiplying by the inverse of $c$on both sides, we get,

$$\begin{gathered} c^{-1}\left(c^2\right)=c^{-1}\left(c\right) \\ \Rightarrow \left(c^{-1}c\right)c=\left(c^{-1}c\right) \\ \Rightarrow ec=e \\ \Rightarrow c=e \end{gathered}$$

Hence, it is proved that, $c=e$.