(e here represents the identity under the operation)
\[ \displaylines{\text{Let }\left(G,\cdot\right)\text{ be a group such that }x^2=e\text{ for all }x\in G.\\ \text{Show that }\left(G,\cdot\right)\text{ is abelian.}} \]
Solution:
\( \displaylines{\text{By hypothesis that }x^2=e\text{ for all }x\in G,\\ aa=bb=e\text{ and }\left(ab\right)\left(ab\right)=e\\ \\ abab=e\\ a\left(abab\right)b=ab\\ aababb=ab\\ ba=ab} \)