# Equivalence relation $g\sim h :\Longleftrightarrow h \in \{g,g^{-1}\}$

Let $(G, \cdot)$ be a group with an Identity element $e$.

(i) A relation on $G$ is defined through $g\sim h :\Longleftrightarrow h \in \{g,g^{-1}\}$. Show that $\sim$ is a equivalence relation and show, that the equivalence class has for $[g]: ~[g]=\{g,g^{-1}\}$

(ii) Let now be $\#G$ even. Show that then $g \in G \setminus\{e\}$ exists with $g^2=e$

I have several problems solving this exercise, one is to write down this correctly To (i): I know i have to show Reflexivity, Symmetry and Transitivity.

Reflexivity: $g\sim g$ is true because $g\sim h :\Longleftrightarrow h \in \{g,g^{-1}\} \Longrightarrow g\sim g$

How to go on with Symmetry and Transitivity?

• I would toss out the abstract notation and say that $h\sim g$ just when $h$ is $g$ or its inverse. Then you need to apply the fact that $h=g^{-1}$ if and only if $g=h^{-1}$ and (equivalently) that $g=(g^{-1})^{-1}$. – Lubin Dec 13 '13 at 19:02
• I changed things like $g\tilde{}h$ to $g\sim h$. That not only looks different but has spacing before and after $\sim$. – Michael Hardy Dec 13 '13 at 20:52

## 1 Answer

For symmetry again go by the definition. Take any two elements $h,\,g\in G$ such that $h\tilde{} g$. Then $g\in \{h,\,h^{-1}\}$. Now there are two cases.

Case I: $g=h$. Then it's obvious that, $g\tilde{}h$

Case II: $g=h^{-1}$. Then $h=g^{-1}\Rightarrow h\in\{g,\,g^{-1}\}\Rightarrow g\tilde{} h$.

Hope you can continue with the the transitivity.

• So the set is the following : $\{(g,h), (g, h^{-1})\}$ If $h \sim g$ and $h^{-1} \sim g \Rightarrow h \sim h^{-1}$? – fear.xD Dec 13 '13 at 19:28
• @fear.xD: No, the relation is $$\{\langle g,g\rangle:g\in G\}\cup\{\langle g,g^{-1}\rangle:g\in G\}\;.$$ – Brian M. Scott Dec 13 '13 at 22:33
• transitivity: $(g,g) \in G$ and $(g, g^{-1}) \in G \Rightarrow (g, g^{-1}) \in G$? – fear.xD Dec 13 '13 at 22:44