Show that in a group $G$ there is exactly one element $g$ for which $g^2 = g$?

Question

Show that in a group $G$ there is exactly one element $g$ for which $g^2 = g$?

What are they essentially asking here? This seems to be a bit general question since its considering all sets that form groups? For the group $(\Bbb{R} \setminus \{0\}, \cdot)$ this would correspond to $1^2=1=e$? Any elaboration/hints on how to show this would be welcome...

Answer 1

It's asking you to show that, no matter the element $g$ in your group $G$, if $g^2=g$, then $g=e$ (since $e^2=e$). So let $g\in G$ such that $g^2=g$. Multiply on, say, the left, by $g^{-1}$. Then

$$\begin{align} e&=g^{-1}g\\ &=g^{-1}g^2\\ &=(g^{-1}g)g\\ &=eg\\ &=g. \end{align}$$

Answer 2

$$g^2=g \Longrightarrow g^{-1}(g^2)=g^{-1}g \Longrightarrow g=e$$

Answer 3

A variant: the simplification rule is valid in groups, so $$g^2(=gg)=g(=ge)\implies g=e.$$