You probably mean "for all " in addition to "for all ".Let H be a subgroup of G with aha(inverse)∈H for all a∈G.

This is easy. To show for all you need to show that for every and every there exists such that . An equivalent equality can be obtained by multiplying both sideson the rightby . Does such actually exist?

For the opposite inclusion, again write the statement to be proved and multiply the equality by on the left. Also, can be renamed into some .