Consider the map $\varphi:\text{GL}(2,\mathbb{Z}_3)\rightarrow \mathbb{Z}_3^*$, where $\varphi(M)=\text{det}(M)$
I have to show that $\varphi$ is a group homomorphism between the groups $(\text{GL}(2,\mathbb{Z}_3),\cdot)$ and $(\mathbb{Z}_3^*,\cdot_3)$. where $\cdot$ is just a matrix multiplication, and $\cdot_3$ is multiplication of integers modulo 3.
My attempt:
I know the definition, hence I choose two 2×2 matrices M and N
$M,N\in \text{GL}(2,\mathbb{Z}_3)$
Then I write up
$\varphi(M\cdot N)=\text{det}(M\cdot N)$
$=\text{det}(M)\cdot_3\text{det}(N)$
This doesn't look right… I'm pretty knew at this 🙁
I tried with a concrete example, but I'm sure that this is wrong…
Thanks in advance
Best Answer
Your solution is correct.
By the definition of a group homomorphism, you have to show that for all $M,N\in\operatorname{GL}(2,\mathbb{Z}_3)$ the equality $\varphi(M\cdot N) = \varphi(M) \cdot \varphi(N)$ is true. Replacing $\varphi$ by its definition this is equivalent to $$\det(M\cdot N) = \det(M) \cdot \det(N),$$ so it boils down to the multiplicativity of the determinant (of matrices over the field $\mathbb{Z}_3$). This is a standard result of linear algebra.
In case that this result has not been covered by your course yet, you can also check it by hand. Write $$M = \begin{pmatrix}a & b\\c & d\end{pmatrix}\quad\text{and}\quad N = \begin{pmatrix}e & f\\g & h\end{pmatrix},$$ and compute both $\det(M\cdot N)$ and $\det(M)\cdot \det(N)$ in terms of $a,b,c,d,e,f,g,h$, to see that indeed $\det(M\cdot N) = \det(M) \cdot \det(N)$ holds.