From Elements of Abstract Algebra question 46$\beta$:
Show that a subgroup H of a group G is normal if and only if ab $\in$ H implies $a^{-1}b^{-1}\in$ H for any elements a,b $\in$ G
Showing left to right is easy: Let H be normal and choose ab $\in$ H. Conjugate by $b^{-1}$ then by $a^{-1}$ and you get $a^{-1}b^{-1}\in$ H
Not sure how to go other direction: Suppose ab $\in$ H implies $a^{-1}b^{-1}\in$ H for any elements a,b $\in$ G and choose $h \in H$ and $g \in G$. Need to show $ghg^{-1} \in H$, but not sure how to proceed. Would appreciate if someone could provide a clue. Can I assume $gh \in H$ and therefore $g^{-1}h^{-1} \in H$? Would that even help?
Note: There are other answers on forum showing that normality implies $ab \in H \therefore ba \in H$. Clearly if you prove this the above is a simple consequence. I was hoping to prove directly.
Suppose that $H$ is not normal. There are are then a $h\in H$ and a $g\in G$ such that $g^{-1}hg\notin H$. Notice that $g.(g^{-1}h)=h\in H$. But $g^{-1}.(g^{-1}h)^{-1}=g^{-1}h^{-1}g=(g^{-1}hg)^{-1}\notin H$.