1. If \(A,B \in M_2(\mathbb{C})\) and \(\lambda \in \mathbb{R}\), then \[\Psi(\lambda A + B) = (\lambda A + B)^\ast = \lambda A^\ast + B^\ast = \lambda \Psi(A) + \Psi(B).\] Thus \(\Psi\) is linear. (This can also be directly checked with explicit matrices.)

2. Since \(\Psi(\Psi(A))=(A^\ast)^\ast=A\), we see that \(\Psi^{\circ 2}\) is the identity. Thus the minimal polynomial of \(\Psi\) divides \(x^2-1\). Thus the eigenvalues can only be \(\pm 1\). The identity matrix \(I_2\) has eigenvalue \(1\) and \(iI_2\) has eigenvalue \(-1\), so they both occur.

3. Explicitly, for \(a,\ldots, h \in \mathbb{R}\), we have \[\Psi : \begin{pmatrix} a+bi & c+di \\ e+fi & g+hi \end{pmatrix} \mapsto \begin{pmatrix} a-bi & e-fi \\ c-di & g-hi \end{pmatrix}.\] Solving \(\Psi(A)=A\) we have a basis for \(E_1\): \[ \begin{pmatrix}1&0\\0&0\end{pmatrix}, \begin{pmatrix}0&0\\0&1\end{pmatrix}, \begin{pmatrix}0&1\\1&0\end{pmatrix}, \begin{pmatrix}0&-i\\i&0\end{pmatrix}.\] Solving \(\Psi(A)=-A\) we have a basis for \(E_{-1}\): \[ \begin{pmatrix}i&0\\0&0\end{pmatrix}, \begin{pmatrix}0&0\\0&i\end{pmatrix}, \begin{pmatrix}0&1\\-1&0\end{pmatrix}, \begin{pmatrix}0&i\\i&0\end{pmatrix}.\]

(You can see these are bases without redoing the computation. They are evidentally linearly independent, and spanning can be checked as follows: \[8 = \operatorname{dim}_{\mathbb{R}} M_2(\mathbb{C}) \ge \operatorname{dim}_{\mathbb{R}} E_1 + \operatorname{dim}_{\mathbb{R}} E_{-1} \ge 4+4=8.)\]

Recall that every element of \(D_{30}\) can be written uniquely as either \(r^i\) or \(sr^i\) for \(0\le i <15\). Every element of the form \(sr^i\) has order \(2\) since \[(sr^i)^2=sr^isr^i=ssr^{-i}r^i=1.\] It remains to consider elements of order \(r^i\), which are precisely the elements of a cyclic subgroup of order \(15\). The possible orders must divide \(15\), so we only consider orders \(1\), \(3\), \(5\) and \(15\). There is only one element of order \(1\). The elements of order \(3\) are of the form \(r^m\) where \(3m \equiv 0 \mod 15\) but \(m \not\equiv 0 \mod 15\); these are \(r^5\) and \(r^{10}\). Similarly, the elements of order \(5\) are \(\{r^3,r^6,r^9,r^{12}\}\). All remaining elements have order \(15\).

Summarizing:

• \(1\) element of order \(1\),
• \(15\) elements of order \(2\),
• \(2\) elements of order \(3\),
• \(4\) elements of order \(5\), and
• \(8\) elements of order \(15\).

Observe that, as a function, \(\sigma(i)\equiv i+1 \pmod{m}\). Thus, \(\sigma^r(i)\equiv i+r\) for all integers \(r\). The set \[ X=\{ 1, \sigma^k(1), \ldots, (\sigma^k)^{m-1} \}\] contains \(m\) distinct elements if and only if \(\sigma\) is an \(m\)-cycle. The set \(X\) does not contain \(m\) distinct elements if and only if \((\sigma^k)^a(1)=(\sigma^k)^b(1)\) for some \(1 \le a < b < m\). This, in turn, is equivalent to \(1+ka \equiv 1+kb \mod{m}\). If \(k\) is relatively prime to \(m\), then this condition is equivalent to \(a \equiv b \mod{m}\) and the elements of \(X\) are distinct. Thus \(\sigma^k\) is a cycle. If \(k\) is not relatively prime to \(m\), then \(lk\) is a multiple of \(m\) for some integer \(0<l<m\). This means that \(1=(\sigma^k)^l(1)\), that \(|X|<m\), and so \(\sigma^k\) is not a cycle.

1. Since the determinant of every matrix in \(G\) is \(1\), we know that \(G\) is a subgroup of \(\operatorname{GL}_3(k)\). Certainly \(G\) is non-empty (e.g. the identity matrix). the product for general elements: \[m\cdot M = \begin{pmatrix}1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix}1 & A & B\\ 0 & 1 & C\\ 0 & 0 & 1\end{pmatrix} = \begin{pmatrix}1 & A+a & B+aC+b\\ 0 & 1 & C+c\\ 0 & 0 & 1\end{pmatrix}.\] show that \(G\) is closed under multiplication. Moreover, every element has an inverse: \[\begin{pmatrix}1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1\end{pmatrix}^{-1} = \begin{pmatrix}1 & -a & -b+ac\\ 0 & 1 & -c\\ 0 & 0 & 1\end{pmatrix}.\] Thus \(G\) is a subgroup of \(\operatorname{GL}_3(k)\); thus, a group.

2. The center \(Z(G)\) is the set of elements of \(G\) that commute with every element of \(G\). Thus, we want to find \(a,b,c \in k\) such that \[M \cdot m \begin{pmatrix}1 & a+A & b+Ac+B\\ 0 & 1 & c+C\\ 0 & 0 & 1\end{pmatrix} = \begin{pmatrix}1 & A+a & B+aC+b\\ 0 & 1 & C+c\\ 0 & 0 & 1\end{pmatrix} = M \cdot m\] for all \(A,B,C \in k\). This reduces immediately to \(aC=cA\), so we require \(a=c=0\). Thus: \[Z(G)= \left\{ \begin{pmatrix}1 & 0 & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}\ \middle|\ b \in k \right\}.\]

3. Set \(Z:=Z(G)\). Observe that \[\begin{pmatrix}1 & a+A & b+Ac+B\\ 0 & 1 & c+C\\ 0 & 0 & 1\end{pmatrix} =\begin{pmatrix}1 & A+a & B+aC+b\\ 0 & 1 & C+c\\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix}1 & 0 & Ac-aC\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}.\] Thus for all \(x,y \in G\), there exists a \(z \in Z\) such that \((xy)=(yx)z\). Thus if \(xZ, yZ \in G/Z\), then \((xZ)(yZ) = xyZ = yxzZ=yxZ=(yZ)(xZ)\). Thus \(G/Z\) is abelian.