Let $G=GL(2,\mathbb{Z}/5\mathbb{Z})$, the general linear group of $2 \times 2 $ matrices with entries from $\mathbb{Z}/5\mathbb{Z}$.
i) List the elements of the centre of $G$.
ii) Compute the order of $G$.
iii) Give an example of an element of $G$ of order $4$, and of an element of $G$ of order $8$.
iv) Let $x$ and $y$ be the elements just constructed. Compute $\det(x)$, $\det(y)$, and $\det(xy)$. What algebraic property of the mapping $\det:G \rightarrow \mathbb{Z}/5\mathbb{Z}$ causes $\det(xy)=\det(x)\det(y)$?
For part i), notice that the only commutative matrices are $rI_n$, where $I_n$ is $n \times n$ identity matrix. Hence, elements of centre of $G$ are $rI_2$, where $r \in \mathbb{Z}_5$
For part ii), since every entry has $5$ choices, we have $5^4$ matrices. But determinant of a matrix will be zero if one row or column is zero or both rows or columns are the same. So, exclude thse matrices , we have $|G|=5^4-5$ (matrix with zero column)$-5$ (matrix with same column and row)
For part iii), notice that $2$ has order $4$ in $\mathbb{Z}_5$, hence consider the matrix $\pmatrix {2 & 0 \\ 0 & 2}$ and it has order $4$. Consider the matrix $\pmatrix {0 & 1 \\ 2 & 0}$ and it has order $8$.
For part iv), let $x = \pmatrix {2 & 0 \\ 0 & 2}$ and $y= \pmatrix {0 & 1 \\ 2 & 0}$. Then $\det(x)=4$ and $\det(y)=-2$ and $\det(xy)=-8$. Then I don't know what is meant by algebraic property here.
Can anyone help me to check my working is valid or not. This is not homework, just an extra exercise for me to prepare for final.
Best Answer
Looks good, idonknow...Nice work.
I do think you undercounted the number of matrices with a zero column: I get $2\cdot5^2 - 1$ (subtracting for the double counting of the zero matrix).
Re: The algebraic property referenced in $(iv)$: I think they might be referring to the fact that the determinant function is a homomorphism between multiplicative groups, which certainly holds for your group. And it is needed to establish that for arbitrary matrices $x, y$ in $\operatorname{GL}(2, \mathbb Z/5\mathbb Z)$, $\det(x, y) = \det(x)\det(y)$. You showed this to be the case with example matrices, but we can use this property of the determinant to generalize and know it must hold for all matrices in your group.