The group of $n \times n$ invertible matrices with entries in $\mathbb{R}$ is denoted $\text{GL}_n(\mathbb{R})$. Similarly, $\text{GL}_n(\mathbb{C})$ denotes the group of $n \times n$ invertible matrices with complex entries. Consider the following sets of matrices:
- $\text{SL}_n(\mathbb{R}) = \{M \in \text{GL}_n(\mathbb{R}) : \det(M) = 1\}$;
- $\text{SL}_n(\mathbb{C}) = \{M \in \text{GL}_n(\mathbb{C}) : \det(M) = 1\}$;
- $\text{O}_n(\mathbb{R}) = \{M \in \text{GL}_n(\mathbb{R}) : MM^\text{T} = M^\text{T}M = I_n\}$;
- $\text{SO}_n(\mathbb{R}) = \{M \in \text{O}_n(\mathbb{R}) : \det(M) = 1\}$;
- $\text{U}_n(\mathbb{C}) = \{M \in \text{GL}_n(\mathbb{C}) : MM^\dagger = M^\dagger M = I_n\}$;
- $\text{SU}_n(\mathbb{C}) = \{M \in \text{U}_n(\mathbb{C}) : \det(M) = 1\}$.
Here $I_n$ stands for the $n \times n$ identity matrix, $M^\text{T}$ is the transpose of $M$, $M^\dagger$ is the conjugate transpose of $M$, and $\det(M)$ denotes the determinant of $M$. Find all possible inclusions among these sets, and prove that in every case the smaller set is a subgroup of the larger one.
The table above has a checkmark when the row element is a subset of the column element. Check the inclusion on a case-by-case basis. For example, if $A\in SU_{n}\left(\mathbb{C}\right)$, then it is in $U_{n}\left(\mathbb{C}\right)$ by definition. However, the diagonal matrix $A=\left(a_{ij}\right)$ with $a_{nn}=i$ and $a_{kk}=1$ for $k<n$ is clearly in $U_{n}\left(\mathbb{C}\right)$ but not in $SU_{n}\left(\mathbb{C}\right)$.
Checking that each is a group is easy. You can use the "one-step subgroup test." $H\neq\emptyset$ is a subgroup of $G$ iff whenever $x,y$ are elements of $H$, $x^{-1}y$ is an element of $H$ (where the group operation is restricted to $H$). Do this on a case-by-case basis. For example, if $A,B\in U_{n}\left(\mathbb{C}\right)$, $A^{\dagger}B\left(A^{\dagger}B\right)^{\dagger}=\left(A^{\dagger}B\right)^{\dagger}A^{\dagger}B=I_{n}$.