Firstly, remember that if and for all , .

For the first part of (a) there are two approaches - either you can apply the fact that any subgroup of index 2 in a group must be normal (why does this apply here?) or you can hit the problem with a stick. That is to say, you need to check that conjugation of every element in by every element of is in . Do you understand how you would do this?

For the second part, you need to find a counter-example. So just try a few things! That is to say, you want to find a such that .

You can actually do parts (b) and (c) together. Prove that this set is equal to and you will have proven that it is a subgroup! To do this, you know that the set has 6 elements (why?) although you do not know if they are distinct. Therefore, prove that all these elements are distinct, and you will have your result! As your group is so small, you can just hit it with a stick (work out what the six elements are). At least, that's what I would do!