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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

Let $A\in GL(2, \mathbb{Z}_n) = \begin{bmatrix}a & b\\ c&d\end{bmatrix}$ and $B \in GL(2, \mathbb{Z}_n) = \begin{bmatrix}e&f\\g&h\end{bmatrix}$. Then the determanent of $A$ is $f(A) = \det(A) = ad-bc$ and the determanent of $B$ is $f(B) = \det(B) = eh-fg$. If we multiply these together, we get $f(A)f(B) = adeh - bceh -adgh + bcgf$. We know that $AB = \begin{bmatrix}ea+gb&fa+hb\\ec+gd&fc+hd\end{bmatrix}$. Then $f(AB) = (ea+gb)(fc+bd) - (ec+gd)(fa+bh)$ is the determinant of $AB$. Some simple algebraic rearrangement of the right-hand side shows that $f(AB) = adeh - bceh -adgh + bcgf$. We also showed that $f(A)f(B) = adeh-bceh-adgh+bcgf$, so we know that $f(AB) = f(A)f(B)$, which means that $f$ is a homomorophism.

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