The identity must be the same, so those matrices have no inverses.
You may prove it like this: If H is a subgroup of G, and x i in H, we have and also which implies that
I am confused about the definition of subgroups.
I am looking at matrices of the form [a 0 ; 0 0] with a not equal to 0. This is a subset of GLn2
I thought this was a subgroup but my book says it's not. I figured that any matrix of that form multiplied by [1 0; 0 0] gives back the matrix, so [1 0; 0 0] is the identity and is in the group. Also, any matrix of that form multiplied [1/a 0; 0 0] gives the previously found identity.
Does the identity of the subgroup need to be the identity of the larger group (in this case [1 0; 0 1]) or the identity that works for the subgroup?
I can see how this would be confusing.
The set you gave is in fact a group in its own right (which is isomorphic to the field from which these matrices take their entries). The problem is that it doesn't get its information from ; in fact, none of the elements in this set are actually in this group.