Show that the groups \(S L(n, \mathbb{R})\) and \(S O(n)\) are closed subgroups of \(G L \cdot(N, \mathbb{R})\), and that \(U(n)\) and \(S U(n)\) are closed subgroups of \(G L(n, C)\). Show furthcrmore that \(S O(n)\) and \(U(n)\) are compact Lie subgroups.

Let's start by defining each of the groups. The special linear group \(SL(n, \mathbb{R})\) comprises of all \(n \times n\) matrices with real entries and determinant equal to 1. Orthogonal group \(SO(n)\) consists of all \(n \times n\) real orthogonal matrices with determinant 1. Unitary group \(U(n)\) is the group of \(n \times n\) unitary matrices, and the special unitary group \(SU(n)\) consists of \(n \times n\) unitary matrices with determinant 1.

A subgroup of \(GL(n, \mathbb{R})\) or \(GL(n, C)\) is closed if it is closed under the operations of the group, which in this case are multiplication and inverse. To prove that \(SL(n, \mathbb{R})\), \(SO(n)\), \(U(n)\) and \(SU(n)\) are closed subgroups we show that multiplying any two matrices in these groups or taking the inverse of a matrix in these groups results in another matrix in the same group. Let's take \(SL(n, \mathbb{R})\) for example. If two matrices from \(SL(n, \mathbb{R})\) are multiplied, the determinant of the product is equal to the product of the determinants (which are both 1), thus the determinant of the product is still 1 and hence belongs to \(SL(n, \mathbb{R})\). The same argument could be made for \(SO(n)\), \(U(n)\), and \(SU(n)\).

A Lie group is compact if it is both closed and bounded. We have already shown that \(SO(n)\) and \(U(n)\) are closed. It remains to show that they are bounded. To check for boundedness of \(SO(n)\), we see that each entry of these matrices is a real number between -1 and 1, thus it is bounded. For \(U(n)\), each entry is a complex number of modulus 1, which is also bounded.Therefore, \(SO(n)\) and \(U(n)\) are both closed and bounded, making them compact Lie subgroups.