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Problem 70 (The Determinant Map Is A Homomorphism)

Let $n$ be an integer greater than 1, and consider the general linear group $\text{GL}(2,\mathbb{Z}_n)$ of two by two invertible matrices mod $n$. Let $f$ be the determinant map $f:\text{GL}(2,\mathbb{Z}_n)\to U(n)$ defined by $f(A)=\det A$, i.e. defined by $$f\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\right)= (ad-bc)\pmod n.$$ Show that $f$ is a homomorphism. Prove this result by direct computation, rather than by claiming it was already shown to be true in a previous course.

After completing this problem, you may assume that the determinant map is always a homomorphism, in other words that $\det(AB)=\det(A)\det(B)$ regardless of the size of the matrix, or the type of coefficients inside.

Solution

From Ben:
  • The letter $n$ below is not part of the mathbb font.
  • You don't want to place restrictions on your choice of $A$ and $B$. You want to pick $A$ and $B$, and then state what this means.
Your current language is placing a restrction (the words "such that" do this). In terms of logic flow, your word choice is actually working in the exact opposite direction of what you want.

Let $A,B \in \text{GL}(2,\mathbb{Z}_n)$. This means that $$A = \begin{bmatrix}a&b\\c&d\end{bmatrix},\text{ and } B = \begin{bmatrix}e&f\\g&h\end{bmatrix}.$$ We will show that $\det(AB) = \det(A)\det(B)$. We know that $$AB = \begin{bmatrix}ae-bg&af-bh\\ce-dg&cf-dh\end{bmatrix}.$$ Thus, we know that $\det(AB) = ((ae + bg)(cf+dh)-(af+bh)(ce+dg)) (\bmod{n}) = (b c f g - a d f g - b c e h + a d e h) (\bmod{n})$.

Now, consider $\det(A)\det(B)$. We know that $\det(A)\det(B) = ((ad-bc) (\bmod{n})) ((eh - fg) (\bmod{n})) (\bmod{n})$. From Problem 25, we know that $((ad-bc) (\bmod{n})) ((eh - fg) (\bmod{n})) (\bmod{n}) = ((ad-bc)(eh - fg)) (\bmod{n})$. Thus, we find that $$\det(AB) = (b c f g - a d f g - b c e h + a d e h) (\bmod{n}) = (b c f g - a d f g - b c e h + a d e h) (\bmod{n}) = \det(A)\det(B)$$.

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