Percisely when the determinant is non-zero.In this case, it consists of the matrices such that a11=a22=a33=1.
Correct.Therefore G has 8 elements.
This might be a mess. Okay I have three ways to show this. The first two are messy. The second is less messy. And the Third is the cleanist but I am not sure if it gonna work.Prove that G is isomorphic with D4.
Okay one way is by trying to define an isomorphism by brute force. The second way is by creating a group table for the group of units and D4 and by brute force trying to find a one-to-one corresponding which forms a homomorphism. The last method is by far the cleanist. But again I am not sure if it is gonna work. Do you have a group table list? I am sure on the internet you can find group tables of certain orders. Okay this is what you do. You find a site that lists all groups up to isomorphism of order 8. Among those are going to be the 3 abelian groups (Z8,Z2xZ4,Z2xZ2xZ2) and 1 non-abelian group (D4). Then knowing that the group of units is not abelian you can say it must be isomorphic to D4 because the table will list all possible non-isomorphic groups. However, if there is another group that is non-abelian then you are in trouble. What you do is by trying to find a structural property that your group of units contains and trying to look at the groups to fails to have that property.
Here is a very nice stite.
Note, the bad news. There are 2 non-abelian groups of order 8 up two isomorphism, D4 and Q (I am supprised I did not catch that). But Q has 6 elements of order 4. While D4 has only 2 elements of order 4. Thus, you need to show that your group of units (which is non-abelian) does not have 6 elements of order 4. Proof Complete.