Group homomorphisms from GL(n,k) to the multiplicative group of k

Posted: March 27, 2024 in Basic Algebra, Groups, Matrices
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Throughout this post, k is a field and k^{\times}:=k \setminus \{0\}, the multiplicative group of k.

We have already seen the general linear group \text{GL}(n,k) and the special linear group \text{SL}(n,k) in this blog several times. The general linear group \text{GL}(n,k) is the (multiplicative) group of all n \times n invertible matrices with entries from k, and the special linear group \text{SL}(n,k) is the subgroup of \text{GL}(n,k) consisting of all matrices with determinant 1.

Since the determinant is multiplicative, the map f: \text{GL}(n,k) \to k^{\times} defined by f(A)=\det A is an onto group homomorphism and \ker f = \text{SL}(n,k). In particular, \text{SL}(n,k) is a normal subgroup of \text{GL}(n,k).

Question. Is f=\det the only group homomorphism \text{GL}(n,k) \to k^{\times}?

Answer. No. For example, the map g: \text{GL}(n,\mathbb{C}) \to \mathbb{C}^{\times} defined by g(A)=\overline{\det A}, where \overline{\det A} is the complex conjugate of \det A, is clearly a group homomorphism. In general, if h: k^{\times} \to k^{\times} is any group homomorphism, then hf: \text{GL}(n,k) \to k^{\times} is also a group homomorphism. In this post, we show that there are no other group homomorphisms \text{GL}(n,k) \to k^{\times}. But first, we need a useful little result from basic group theory.

Lemma. Let G_1,G_2 be groups, and let f,g: G_1 \to G_2 be group homomorphisms. If f is onto and \ker f \subseteq \ker g, then there exists a group homomorphism h: G_2 \to G_2 such that g=hf.

Proof. You can prove that directly by showing that the map h defined by h(f(x))=g(x) is a well-defined group homomorphism from G_2 to G_2. Here is another way. Let K_1:=\ker f and K_2:=\ker g. Since f is onto, the map \alpha : G_1/K_1 \to G_2 defined by \alpha(xK_1)=f(x), \ x \in G_1, is a group isomorphism. Since K_1 \subseteq K_2, the map \beta : G_1/K_1 \to G_1/K_2 defined by \beta(xK_1)=xK_2, \ x \in G_1, is a well-defined group homomorphism. Finally, we have the injective group homomorphism \gamma : G_1/K_2 \to G_2 defined by \gamma(xK_2)=g(x), \ x \in G_1. Now, let h:=\gamma \beta \alpha^{-1}. Then h: G_2 \to G_2 is a group homomorphism and for all x \in G_1,

hf(x)=\gamma \beta \alpha^{-1}(f(x))=\gamma \beta(xK_1)=\gamma(xK_2)=g(x). \ \Box

Theorem. A map g: \text{GL}(n,k) \to k^{\times} is a group homomorphism if and only if g=hf, where f: \text{GL}(n,k) \to k^{\times} is defined by f(A)=\det A, \ A \in \text{GL}(n,k), and h: k^{\times} \to k^{\times} is any group homomorphism.

Proof. If h: k^{\times} \to k^{\times} is a group homomorphism, then obviously g=hf: \text{GL}(n,k) \to k^{\times} is a group homomorphism. Conversely, suppose now that g: \text{GL}(n,k) \to k^{\times} is a group homomorphism. We consider two cases.

Case 1: |k|=2. In this case, k^{\times}=(1) and so the only group homomorphism \text{GL}(n,k) \to k^{\times} or k^{\times} \to k^{\times} is the trivial one. Thus f,g,h in this case are all trivial maps.

Case 2: |k| > 2. Let A,B \in \text{GL}(n,k). Then, since the image of g is an abelian group,

g(ABA^{-1}B^{-1})=g(A)g(B)g(A^{-1})g(B^{-1})=g(A)(g(A))^{-1}g(B)(g(B))^{-1}=1

and so ABA^{-1}B^{-1} \in \ker g. Therefore the commutator subgroup of \text{GL}(n,k) is contained in \ker g. On the other hand, by this post, the commutator subgroup of \text{GL}(n,k) is \text{SL}(n,k)=\ker f. So \ker f \subseteq \ker g and the result now follows from the Lemma. \ \Box

Note. For some variations and generalizations of the Theorem, see this paper.

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