About the closure of the subset of n'th roots of the identity matrix. I'm working with the square roots of the identity matrix and it seems that the only thing that prevents this subset of the general linear group from being a subgroup is closure. Since the identity itself is in this subset and every element is it's own inverse, I wonder how the smallest subgroup of the general linear group that contains this subset looks like and how to generalize this with n'th roots instead of just square roots.
(New guy over here, sorry if I made any mistakes regarding the rules for asking and such)
Edit: I'm working over the complex numbers, but I guess it could be over any field.
 A: You say "the" general linear group but, in general, the answers to such questions can depend on the dimension or the field you are working over. (Although not much in this case.)
The subgroup you are considering is clearly a normal subgroup of the general linear group, as the generating set is closed under conjugation. (In fact, it is clearly a characteristic subgroup.) 
In most cases, "the" general linear groups has very few normal subgroups. They are either "small" and contained in the center (and consist of scalar matrices), or "large" and contain the special linear group. 
(There are some exceptions to this statement, for example $GL(2,3)$.)
Now, the subgroup you are considering clearly has non-scalar matrices (like some permutation matrices of order $2$), so it must contain the special linear group. In particular, it is completely determined by the set of determinants of matrices represented. If we are working over a field, a matrix of order $2$ must have determinant $\pm 1$. You can take a matrix with $1$ on the diagonal, except one $-1$, to show that you do have some matrices of determinant $-1$. So the group is exactly the set of matrices of determinant $\pm 1$. 
