the inverse of -1 is -1 and is in the set

1 times -1 = -1 which is in the set which shows closure

but i thought you said 1 couldnt be in the subset?
No - the subset cannot be precisely $\{1\}$ if it is trivial, but 1 must be in it.

2. Originally Posted by Swlabr
No - the subset cannot be precisely $\{1\}$ if it is trivial, but 1 must be in it.
so the non trivial subgroups are

{1,-1}
{1,i}
{1, -i}

{1,-1,i}
{1,-1,-i}

so the non trivial subgroups are

{1,-1}
{1,i}
{1, -i}

{1,-1,i}
{1,-1,-i}
No. Remember what the theorem you are trying to show says. Which of your sets don't adhere to what it says?

The problem with some of your given sets is closure: $a, b \in H \Rightarrow ab \in H$.

4. Originally Posted by Swlabr
No. Remember what the theorem you are trying to show says. Which of your sets don't adhere to what it says?

The problem with some of your given sets is closure: $a, b \in H \Rightarrow ab \in H$.
bottom two because -1 * 1 * i = -i which is not in the subgroup

-1 * 1 * -i = i which is not in the subgroup