Matrix Groups: GL, SL, O, and U

We use the subgroup criterion technique: to show a nonempty subset $H$ of a group $G$ is a subgroup, it suffices to show $H$ is closed under the group operation and under taking inverses (closure under the operation, together with inverses, forces the identity to lie in $H$ as well, since for any $h\in H$, $h^{-1}\in H$ by assumption and then $hh^{-1}=e\in H$ by closure). We apply this after first establishing $GL_n(F)$ itself is a group, since all other claims are subgroup claims relative to it.

Step 1: $M_n(F)$ under multiplication is a monoid, and $GL_n(F)$ is its group of units. Matrix multiplication on $M_n(F)$ is associative: for $A,B,C \in M_n(F)$, the $(i,j)$ entry of $(AB)C$ and of $A(BC)$ both equal

by direct expansion of the double sum defining matrix multiplication (this is a finite algebraic identity in the field $F$, using only commutativity and associativity of $+,\times$ in $F$ and the distributive law, and holds entrywise). The identity matrix $I$ satisfies $IA = AI = A$ for all $A$, by direct computation of matrix entries ($I_{ik} = \delta_{ik}$, so $(IA)_{ij} = \sum_k \delta_{ik}A_{kj} = A_{ij}$, and symmetrically on the other side). Thus $(M_n(F), \cdot)$ is a monoid with identity $I$. Now recall the basic linear-algebra fact (Cramer/adjugate theory over a field): a matrix $A \in M_n(F)$ has a two-sided multiplicative inverse in $M_n(F)$ if and only if $\det(A) \ne 0$, and in that case the inverse is unique and is given by

where $\mathrm{adj}(A)$ is the adjugate (transpose of the cofactor matrix), an expression built from $F$-polynomial combinations of the entries of $A$, hence itself lying in $M_n(F)$. We take this classical fact from linear algebra as known (it does not depend on group theory, so invoking it here is not circular). Consequently $GL_n(F)$, defined as $\{A : \det A \ne 0\}$, is precisely the set of units of the monoid $M_n(F)$. It is a general fact, verified directly, that the set of units of any monoid forms a group under the monoid operation: it is closed under multiplication (if $A, B$ are units with inverses $A^{-1}, B^{-1}$, then $B^{-1}A^{-1}$ satisfies $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1}=I$ and similarly on the other side, so $AB$ is a unit with inverse $B^{-1}A^{-1}$), contains the identity ($I$ is its own inverse), and is closed under inversion (if $A$ is a unit with inverse $A^{-1}$, then $A^{-1}$ is a unit with inverse $A$). Associativity is inherited from $M_n(F)$. Hence $GL_n(F)$ is a group. This proves claim (1).

Step 2: $\det$ is multiplicative, so $SL_n(F)$ is a subgroup. The determinant satisfies $\det(AB) = \det(A)\det(B)$ for all $A, B \in M_n(F)$; this is a standard identity of linear algebra (provable via the Leibniz expansion of the determinant as a signed sum over permutations, or via elementary row operations), independent of group theory, and we invoke it as known. Since $\det(I) = 1 \ne 0$, we have $I \in GL_n(F)$, and clearly $\det(I) = 1$ so $I \in SL_n(F)$; thus $SL_n(F)$ is nonempty. Closure under multiplication: if $A, B \in SL_n(F)$, then $\det(A) = \det(B) = 1$, so $\det(AB) = \det(A)\det(B) = 1\cdot 1 = 1$, and since $1 \ne 0$, also $AB \in GL_n(F)$. Hence $AB \in SL_n(F)$. Closure under inversion: if $A \in SL_n(F)$, then $A \in GL_n(F)$, so $A^{-1}$ exists in $GL_n(F)$, and from $AA^{-1} = I$ we get $\det(A)\det(A^{-1}) = \det(I) = 1$; since $\det(A) = 1$, this forces $\det(A^{-1}) = 1$. Hence $A^{-1} \in SL_n(F)$. By the subgroup criterion, $SL_n(F)$ is a subgroup of $GL_n(F)$, proving claim (2).

Step 3: $O(n)$ is a subgroup of $GL_n(\mathbb{R})$. First we check $O(n) \subseteq GL_n(\mathbb{R})$: if $A^TA = I$, then taking determinants of both sides (using $\det(A^T) = \det(A)$, another standard fact of linear algebra) gives $\det(A)^2 = \det(A^T)\det(A) = \det(A^TA) = \det(I) = 1$, so $\det(A) = \pm 1 \ne 0$, hence $A \in GL_n(\mathbb{R})$. This simultaneously proves the claimed fact that every element of $O(n)$ has determinant $\pm1$. Nonemptiness: $I^T I = I \cdot I = I$, so $I \in O(n)$. Closure under multiplication: suppose $A, B \in O(n)$, so $A^TA = I$ and $B^TB=I$. Using $(AB)^T = B^TA^T$ (a standard identity: transpose reverses order of a product, provable entrywise, $((AB)^T)_{ij} = (AB)_{ji} = \sum_k A_{jk}B_{ki} = \sum_k (B^T)_{ik}(A^T)_{kj} = (B^TA^T)_{ij}$), we compute

Hence $AB \in O(n)$. Closure under inversion: let $A \in O(n)$. Since $A \in GL_n(\mathbb{R})$ by the above, $A^{-1}$ exists. From $A^TA = I$, multiply both sides on the left by $(A^T)^{-1}$ and on the right appropriately: actually more directly, $A^TA = I$ means $A^T$ is a left inverse of $A$; since $A$ is invertible with unique two-sided inverse $A^{-1}$, a left inverse must equal the two-sided inverse, so $A^T = A^{-1}$. Then

using $(A^T)^T = A$. Hence $A^{-1} \in O(n)$. By the subgroup criterion, $O(n)$ is a subgroup of $GL_n(\mathbb{R})$, proving claim (3).

Step 4: $U(n)$ is a subgroup of $GL_n(\mathbb{C})$, by an identical argument with conjugate transpose. The conjugate transpose satisfies $(AB)^* = B^*A^*$ (since $^*$ is transpose composed with entrywise complex conjugation, and both transpose-reversal and conjugation of a product being the product of conjugates in either order are standard facts: $\overline{AB} = \overline{A}\,\overline{B}$ entrywise since complex conjugation is a ring homomorphism of $\mathbb{C}$, combined with the transpose-reversal identity above) and $(A^*)^* = A$. If $A^*A = I$, then taking determinants: $\det(A^*)\det(A) = \det(I) = 1$. Since $\det(A^*) = \overline{\det(A)}$ (conjugating every entry of $A$ before taking the determinant, a polynomial in the entries with integer coefficients, conjugates the determinant, as complex conjugation is a field automorphism of $\mathbb{C}$ fixing $\mathbb{Z}$), we get $\overline{\det(A)}\det(A) = |\det(A)|^2 = 1$, so $|\det(A)| = 1 \ne 0$; in particular $A \in GL_n(\mathbb{C})$. This proves the claimed fact that every unitary matrix has determinant of absolute value $1$. Nonemptiness: $I^*I = I$, so $I \in U(n)$. Closure under multiplication: for $A, B \in U(n)$,

so $AB \in U(n)$. Closure under inversion: as before, $A^*A = I$ with $A$ invertible forces $A^* = A^{-1}$ (a left inverse of an invertible matrix is the two-sided inverse), and then

so $A^{-1} \in U(n)$. By the subgroup criterion, $U(n)$ is a subgroup of $GL_n(\mathbb{C})$, proving claim (4). All four claims are established, and associativity throughout is inherited automatically from the ambient monoid $M_n(F)$, so no separate verification of associativity is needed for the subgroups. $\blacksquare$