*Algebra*

## Miscellaneous problems for Chapter 7

Problem 7.M.1 Every element can be written as some product of alternating \(x\)'s and \(y\)'s: we never need to use \(x^{-1}\) or \(y^{-1}\) since these are just \(x\) and \(y\) respectively, and any two adjacent \(x\)'s or \(y\)'s can be cancelled. The analysis proceeds by cases:

- Case 1: The group is infinite. The none of the nontrivial alternating words can be equal to the identity, because if for example \(xyxy = 1\), then conjugation by \(y\) implies that \(yxyx = 1\) as well, and all longer words would then just be equal to one of the shorter words. Furthermore, two different alternating words must represent different elements: if \(W_1 = W_2\) then \(W_1 W_2^{-1} = 1\) and we could derive a contradiction. So \(G\) consists precisely of the elements \(1, x, y, xy, yx, xyx, yxy, \ldots\), which are all distinct; because the multiplication rule is known, there is at most one isomorphism class of such groups. This group can be explicitly constructed using the generators \[ x = \begin{pmatrix} 0 & t \\ t^{-1} & 0 \end{pmatrix}, \qquad y = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \] where \(t\) is a formal symbol (but can be replaced by any nonzero number that isn't a root of unity); it consists of all diagonal matrices with the elements \(t^n\) and \(t^{-n}\) for each \(n \in \mathbb{Z}\) as well as all antidiagonal matrices of this form.
- Case 2: The group is finite. Then some alternating word has to equal the identity. Say the shortest alternating word that equals the identity has length \(n\). If \(n\) is odd, then the first and last factor are the same, for example, \(xyxyx = 1\). Conjugation first by \(x\) and then by \(y\) yields \(x = 1\), a contradiction. This occurs for all odd \(n\), therefore \(n\) must be even. Write \(z = xy\); clearly \(z^{-1} = yx\). Write \(n = 2m\). By assumption, the element \(z\) has order \(m\); furthermore, it is easy to see by direct conjugation with \(x\) and \(y\) that \(\langle z \rangle\) is a normal subgroup of \(G\). Now each alternating word of odd length that starts with \(x\) can be written as \(xz^{-i}\) where \(i\) is half of one less than the length of the word. If an alternating word \(W\) of odd length starts with \(y\), say \(W = yz^i\), then \(xW = xyz^i = z^{i+1}\) so that \(W = xz^{i+1}\). This establishes that \(G = \langle x \rangle \langle z \rangle\). Note that \(x \notin \langle z \rangle\) because if \(x = z^i\) for some \(i\) then multiplying both sides by \(x\) would give an odd-length word equal to 1, which we already argued can't happen. By Theorem 2.11.4, \(G = \langle x\rangle \langle z \rangle\), and by direct calculation, \(xz^ix^{-1} = z^{-i}\). So \(G \cong D_m\). The generators \(x\) and \(y\) could be two reflections with the angle \(\pi/m\) between their lines.

Problem 7.M.2 We can do this by direct calculation. Suppose \(g = x^i\) for some \(i\). Then \(HgH\) consists of the elements \(1x^i 1, rx^i 1, 1x^i r, rx^i r\), which respectively equal \(x^i, x^{-i}r, x^i r, x^{-i}\). If \(n \divides 2i\), then this is really only two elements \(x^i, x^i r\); otherwise it is four elements. Since these double cosets already contain among them all elements of the form \(x^j r\) as well, we are done.

Problem 7.M.3(a) Let \(H\backslash G/H\) denote the set of double cosets of \(H\) in \(G\) and let \(\mathcal{O}\) denote the orbits of \(S \times S\). Define \(f: H\backslash G/H \to \mathcal{O}\) as follows: \(f(HgH)\) is the orbit containing the element \((s_0, gs_0)\). We will first show that this is well-defined. Suppose \(g, g' \in G\) with \(HgH = Hg'H\). Then \(g' \in HgH\), so write \(g' = h_1 g h_2\) for some \(h_1, h_2 \in H\). Then \((s_0, g's_0) = (s_0, h_1 g h_2 s_0) = (s_0, h_1 g s_0) = h_1 (s_0, gs_0)\), so \((s_0, g's_0)\) belongs to the same orbit as \(s_0, gs_0\). So \(f\) is well-defined. Next, since \(G\) acts transitively on \(S\), the image of \(f\) contains all orbits that contain an element of the form \((s_0, s)\) where \(s \in S\), and again, since \(G\) acts transitively on \(S\), every orbit of \(S \times S\) under \(G\) contains some element of this form; therefore \(f\) is surjective. Finally, we will show that \(f\) is injective too. Suppose \(f(Hg_1 H) = f(Hg_2 H)\). Then \((s_0, g_2 s_0) = g'(s_0, g_1 s_0)\) for some \(g' \in G\). But then \(g' s_0 = s_0\), so \(g' \in H\). Write \(h_1 = g'\). Then \((s_0, g_2 s_0) = (s_0, h_1 g_1 s_0)\). Since \(g_2 s_0 = h_1 g_1 s_0\), it follows that \(s_0 = g_2^{-1} h_1 g_1 s_0\). This implies \(g_2^{-1} h_1 g_1 = h_2\) for some \(h_2 \in H\), whence \(g_2 = h_1 g_1 h_2^{-1}\). That is, \(g_2 \in Hg_1 H\), so the two double cosets \(Hg_1 H\) and \(Hg_2 H\) are the same. This establishes that \(f\) is injective. So \(f\) is a bijection.

Problem 7.M.4 \(H\) is not necessarily normal in \(G\). For example, let \(G\) be \(A_4\), let \(K\) be the normal Klein four-subgroup of \(G\), and let \(H\) be a subgroup of \(K\) of order two. Then \(H\) is not normal in \(G\). For example, \(\{1, (1\ 2)(3\ 4)\}\) is not normal in \(A_4\).

Problem 7.M.5

- Since \(\ker \pi = N\), it is obvious that \(\ker \pi|_H\) is just \(H \cap N\), and \(\ker \pi|_{HN}\) is just \(HN \cap N\), which is just \(N\) since \(N \subseteq HN\).
- By the first isomorphism theorem, \(H/(H \cap N) = H/\ker \pi|_H \cong \pi(H)\) and \(HN/N = HN/\ker \pi|_{HN} \cong \pi(HN)\). But \(\pi(H) = \pi(HN)\) since \(N = \ker \pi\). Therefore \(H/(H \cap N) \cong HN/N\).

Problem 7.M.6

- For every \([h N] \in \overline{H}\) and \([gN] \in \overline{G}\), we have that \([gN][hN][gN]^{-1} = [ghg^{-1}N] = [h'N]\) where \(h' = ghg^{-1}\). Since \(H\) is normal in \(G\), \(h' \in H\), and \([h'N] \in \overline{H}\). Therefore \(\overline{H}\) is normal in \(\overline{G}\).
- Let \(\pi_1 : G \to \overline{G}\) and \(\pi_2 : \overline{G} \to \overline{G}/\overline{H}\) be the canonical homomorphisms. If \(h \in H\), then \(\pi_2(\pi_1(h)) = \pi_2([hN]) = 1\) since \([hN] \in \overline{H}\). Conversely, if \(g \in G\) satisfies \(\pi_2(\pi_1(g)) = 1\), then \(\pi_1(g) \in \ker \pi_2 = \{[hN] \mid h \in H\}\), that is, \(\pi_1(g) = [hN]\) for some \(h \in H\). But \(\pi_1(g) = [gN] = [hN]\) so \(g \in hN\) implying \(g \in H\) since \(N \subseteq H\). So \(\ker(\pi_2 \circ \pi_1) = H\). By the first isomorphism theorem, \(\overline{G}/\overline{H} = \im(\pi_2 \circ \pi_1) \cong G/\ker(\pi_2 \circ \pi_1) = G/H\).

Problem 7.M.7

- \(p_1\) can be written as a product of cycles involving only elements in
\(U_1\),
*i.e.,*\(p_1 = C_1 C_2 \ldots C_m\). Likewise \(p_2 = D_1 D_2 \ldots D_n\) where all the \(C\)'s and \(D\)'s are disjoint. Therefore \(p_1 p_2 = p_2 p_1 = C_1 C_2 \ldots C_m D_1 D_2 \ldots D_n\). - Let \(x\) be the single element in \(U_1 \cap U_2\). Let \(C_1\) be the cycle in \(p_1\) that contains \(x\) and let \(D_1\) be the cycle in \(p_2\) that contains \(x\). Then \(p_1 p_2 = C_1 D_1 (C_2 C_3 \ldots C_m D_2 D_3 \ldots D_n)\) and \(p_1^{-1} p_2^{-1} = C_1^{-1} D_1^{-1} (C_2 C_3 \ldots C_m D_2 D_3 \ldots D_n)^{-1}\) and the commutator is simply that of \(C_1\) and \(D_1\). Write \(C_1 = (x\ a_1\ a_2 \ldots a_q), D_1 = (x\ b_1\ b_2 \ldots b_r)\). Then direct calculation shows \begin{align*} C_1 D_1 &= (x\ b_1\ b_2 \ldots b_r\ a_1\ a_2 \ldots a_q) \\ C_1^{-1} D_1^{-1} &= (x\ b_r\ b_{r-1} \ldots b_1\ a_q\ a_{q-1} \ldots a_1) \\ [C_1, D_1] &= (x\ a_1\ b_1) \end{align*}

Problem 7.M.8 Let \(C\) be a left coset of \(H\) in \(G\). Then there is some \(g\) such that \(C = gH\), and the set \(C^{-1}\) consisting of the inverses of elements of \(C\) is equal to \(H^{-1}g^{-1} = Hg^{-1}\), which is a right coset. It is easy to see that the inverse of a right coset is similarly a left coset. This establishes a bijection between the set of left cosets and the set of right cosets.

Problem 7.M.9 Suppose \(x\) is conjugate to \(x^{-1}\). Let \(C\) denote the conjugacy class of \(x\). Whenever \(C\) contains the element \(y = gxg^{-1}\), it also contains the element \(g^{-1}x^{-1}g\), which is the inverse of \(y\). Since the group is of odd order, no element other than the identity equals its own inverse. Obviously \(C\) doesn't contain the identity, so the elements of \(C\) can all be placed into pairs with their inverses. Therefore \(|C|\) is even. A group of odd order can't contain a conjugacy class of even cardinality, so a contradiction has been reached.

Problem 7.M.10 Let \(S^g\) denote the number of elements of \(S\) that are fixed by \(g \in G\). Since \(G\) acts transitively, by Burnside's lemma, \(|G| = \sum_{g\in G} S^g\). But \(S^1 \ge 2\), so \(\sum_{g \in G \setminus \{1\}} S^g \le |G| - 2\). By the pigeonhole principle, at least one of the \(|G| - 1\) summands must be zero.

Problem 7.M.11 Most of the work involves the matrices of order two with determinant −1. The following two Lemmas involve such matrices:

*Lemma 1:* Suppose
\[
M = \begin{pmatrix} a & b \\ c & -a \end{pmatrix}
\]
where \(\det M = -1\). Then \(M\), regarded as an element of
\(GL_2(\mathbb{Z})\), is conjugate to one of the following matrices:
\[
J = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad
K = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
\]
where \(x \in \mathbb{Z}\).

*Proof:* If \(a \lt 0\), replace \(M\) by its conjugate \(JMJ^{-1}\).
We will therefore assume from this point on that \(a \ge 0\). Let
\[
E = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \qquad
E' = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}
\]
and consider the following conjugates of \(M\):
\begin{alignat*}{2}
M_1 &= E^{-1} M E
&&=
\begin{pmatrix} a - c & 2a + b - c \\ c & -a + c \end{pmatrix} \\
M_2 &= E M E^{-1}
&&=
\begin{pmatrix} a + c & -2a + b - c \\ c & -a - c \end{pmatrix} \\
M_3 &= E' M E'^{-1}
&&=
\begin{pmatrix} a - b & b \\ 2a - b + c & -a + b \end{pmatrix} \\
M_4 &= E'^{-1} M E'
&&=
\begin{pmatrix} a + b & b \\ -2a - b + c & -a - b \end{pmatrix}
\end{alignat*}
If \(c \ne 0\) and \(|c| \le a\), then at least one of \(M_1\) and \(M_2\)
has a top-right entry which is nonnegative and strictly less than \(a\).
If \(b \ne 0\) and \(|b| \le a\), then at least one of \(M_3\) and \(M_4\) has
a top-right entry which is nonnegative and strictly less than \(a\).
If any of
these four conjugates has this property, call it \(M'\). Note that \(M'\) has
the same properties as \(M\), that is, \(M'_{11} \ge 0\), \(M'_{22} =
-M'_{11}\), and \(\det M' = -1\). Let us suppose that
repeated conjugation is used to reduce the absolute value of the diagonal
entries, giving a sequence of conjugates \(M, M', M'', \ldots, M_0\) where
\(M_0\) is a matrix for which no further reduction is possible. Write
\[
M_0 = \begin{pmatrix} a_0 & b_0 \\ c_0 & -a_0 \end{pmatrix}
\]
where \(a_0 \ge 0\) and \(\det M_0 = -1\). One of the following must now
hold:

- Case 1: Both \(b_0\) and \(c_0\) are nonzero. Then it must be that \(|b_0| \ge a_0 + 1\) and \(|c_0| \ge a_0 + 1\), otherwise a further reduction would be possible. Then \(|b_0c_0| \ge a_0^2 + 2a_0 + 1\), so \(1 = |\det M_0| = |-a_0^2 - b_0c_0| \ge 2a_0 + 1\) implying \(a_0 = 0\). Since \(b_0, c_0 \in \mathbb{Z}\), the condition \(\det M_0 = -1\) implies that either \(M_0 = J\) or \(M_0 = -J\). If \(M_0 = J\) then \(M\) is conjugate to \(J\). If \(M_0 = -J\) then \(KM_0K^{-1} = J\), and again \(M\) is conjugate to \(J\).
- Case 2: \(b_0 = 0\) and \(c_0\) is even. Then the condition \(\det M_0 = -1\) implies \(a_0 = 1\). Then \(E'^{-c_0/2}M_0E'^{c_0/2} = K\). Therefore \(M\) is conjugate to \(K\).
- Case 3: \(c_0 = 0\) and \(b_0\) is even. This is similar to Case 2. Here \(E^{b_0/2} M_0 E^{-b_0/2} = K\). Again \(M\) is conjugate to \(K\).
- Case 4: \(b_0 = 0\) and \(c_0\) is odd. As in Case 2, repeated conjugation by \(E'\) or \(E'^{-1}\) can be used to change the value of \(c_0\) by 2 at a time, yielding the conjugate \[ M_1 = \begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix} \] Then \(E^{-1} M_1 E = J\), so \(M\) is conjugate to \(J\).
- Case 5: \(c_0 = 0\) and \(b_0\) is odd. As in Case 3, repeated conjugation by \(E\) or \(E^{-1}\) can be used to obtain a conjugate \[ M_1 = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix} \] Then \(E' M_1 E'^{-1} = J\), so \(M\) is conjugate to \(J\).

*Lemma 2:* Say a matrix \(M\) is of *K-type* if \(M \in
GL_2(\mathbb{Z})\), the two diagonal entries are odd, and the other two entries
are odd. All conjugates of a K-type matrix are of K-type.

*Proof:* We showed in
Problem 2.M.14 that the matrices \(E, E'\) generate
\(SL_2(\mathbb{Z})\). This implies that \(E, E'\) together with a matrix of
determinant −1, such as \(J\), are sufficient to generate
\(GL_2(\mathbb{Z})\). From the explicit expressions given in
the proof of Lemma 1, observe that the conjugate of a K-type matrix by any of
\(E, E^{-1}, E', E'^{-1}\) is also of K-type. The conjugate of a K-type matrix
by \(J\) is also of K-type (note that \(J\) is its own inverse). Since
\(E, E', J\) generate \(GL_2(\mathbb{Z})\), the desired result follows.

*Corollary:* The matrices \(J\) and \(K\) are not conjugate to each
other in \(GL_2(\mathbb{Z})\).

We are now ready to prove the main result. Suppose \(M\) is of order 2 in \(GL_2(\mathbb{Z})\). If \(\det M = 1\) then \begin{align*} M &= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \\ M^{-1} &= \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \end{align*} Now \(M = M^{-1}\) so \(a = d, b = 0, c = 0\), and the condition \(\det M = 1\) then gives \(a = \pm 1\). So \(M = \pm I\).

If on the other hand \(\det M = -1\) then \[ M^{-1} = \begin{pmatrix} -d & b \\ c & -a \end{pmatrix} \] whence \(a = -d\) and \(M\) is of the form described in Lemma 1. This implies that \(M\) belongs to the conjugacy class of either \(J\) or \(K\). By Lemma 2, if \(M\) is of K-type then it's conjugate to \(K\), otherwise it's conjugate to \(J\).

Using the fact that scalar matrices are similar only to themselves, and that conjugation preserves the determinant, we conclude that there are four conjugacy classes of elements of \(GL_2(\mathbb{Z})\) of order 2:

- The conjugacy class consisting only of \(I\);
- The conjugacy class consisting only of \(-I\);
- The conjugacy class of elements of order 2 with determinant −1 that are of K-type;
- The conjugacy class of elements of order 2 with determinant −1 that are not of K-type.

Problem 7.M.12

- Let \(M_a = \begin{pmatrix} 0 & -1 \\ 1 & a\end{pmatrix}\). Suppose
\begin{equation}
M_a \begin{pmatrix} b & c \\ d & e \end{pmatrix} =
\begin{pmatrix} b & c \\ d & e \end{pmatrix} M_a
\label{eqn:M_a_comm}
\end{equation}
By multiplying out both sides, we arrive at the system of equations
\begin{align*}
-d &= c \\
-e &= -b + ac \\
b + ad &= e \\
c + ae &= -d + ae
\end{align*}
The first and second equations together imply the third and fourth. It is clear
that choosing values of \(b\) and \(d\) is sufficient to determine unique
values for \(c\) and \(e\), but the condition \(be - cd = 1\) must be satisfied
for the matrix to lie in \(SL_2(\mathbb{F}_5)\). Substituting the expressions
for \(c\) and \(e\) in terms of \(b\) and \(d\) yields
\begin{equation*}
b^2 + abd + d^2 = 1
\end{equation*}
For all \(a\), the solutions \((b, d) = (0, \pm 1), (\pm 1, 0)\) exist,
corresponding to \(\pm M_a\) and \(\pm I\). For either \(b\) or \(d\) zero, it
is easy to see that no other solutions exist for any \(a\).
If \(b\) and \(d\) are nonzero, then \(bd\) is nonzero and there is exactly one
\(a\) for which \(abd = 1 - b^2 - d^2\), giving the solutions:
- \(a = 0\): no other solutions.
- \(a = 1\): \((1, 4), (4, 1)\)
- \(a = 2\): \((1, 3), (2, 2), (2, 4), (3, 1), (3, 3), (4, 2)\)
- \(a = 3\): \((1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3)\)
- \(a = 4\): \((1, 1), (4, 4)\)

- \(a = 0\): The matrix \(M\) has order 4, so the centralizer \(Z(M)\), which is known to have order 4, must simply be \(\langle M\rangle \cong C_4\).
- \(a = 1\): The matrix \(M\) has order 6 and \(Z(M)\) has order 6, therefore \(Z(M) = \langle M \rangle \cong C_6\).
- \(a = 2\): The matrix \(-M\) has order 10 and \(Z(M)\) has order 10, therefore \(Z(M) = \langle -M \rangle \cong C_{10}\).
- \(a = 3\): The matrix \(M\) has order 10 and \(Z(M)\) is known to have order 10, therefore \(Z(M) = \langle M \rangle \cong C_{10}\).
- \(a = 4\): The matrix \(-M\) has order 6 and is known to belong to \(Z(M)\), which has order 4, so \(Z(M) = \langle -M \rangle \cong C_6\).

The order of \(SL_2(\mathbb{F}_5)\) is \(\frac{1}{4}(5^2 - 1)(5^2 - 5) = 120\). The result of part (a) implies the existence of conjugacy classes of orders 30, 20, 12, 12, and 20, which are necessarily disjoint since conjugation preserves the trace. The matrices \(I\) and \(-I\) belong to singleton conjugacy classes. This leaves 24 elements unaccounted for.

We know that the conjugacy class of \(M_2\) contains only elements of trace 2, yet case analysis shows that there are 25 elements in \(SL_2(\mathbb{F}_p)\) of trace 2, of which 13 have been accounted for, namely the 12 belonging to the conjugacy class of \(M_2\), and the identity. We therefore look for 12 additional elements of trace 2. We will make use of the following Lemmata:

*Lemma 1:*Let \(a, b \in \mathbb{F}_p \setminus \{0\}\). Then the following matrices in \(SL_2(\mathbb{F}_p)\), \[ N_a = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}, \qquad N_b = \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix} \] are conjugate to each other iff \(a/b\) is the square of some element of \(\mathbb{F}_p\).*Proof:*Expand out the equation \(AN_a = N_bA\) and use the condition \(\det N = 1\). The details are left as an exercise.*Lemma 2:*The centralizer of the matrix \(N_a\) defined in Lemma 1 is the set of matrices of the form \(\begin{pmatrix} x & y \\ 0 & x \end{pmatrix}\) with determinant 1. The number of such matrices is \(2p\) whenever \(p \ne 2\).*Proof:*Expand out the equation \(AN_a = N_aA\) and solve. Once the form of the matrices in \(Z(N_a)\) has been found, observe that \(y\) can be chosen arbitrarily while \(x\) can be \(\pm 1\); this proves that there are \(2p\) such matrices. The details are left as an exercise.According to Lemma 1, the matrices \(N_1\) and \(N_2\) are not conjugate in \(SL_2(\mathbb{F}_p)\) but both have trace 2. According to Lemma 2, both have a centralizer of order 10 and therefore a conjugacy class of size 12. At least one of the two conjugacy classes is not that of \(M_2\) (we don't bother to determine which; it doesn't matter), so it is a conjugacy class of size 12 that hasn't previously been accounted for. The matrices \(-N_1\) and \(-N_2\) are also not conjugate to each other and each has trace 3; this gives us an additional conjugacy class of size 12 that doesn't contain \(M_3\). We conclude that the class equation of \(SL_2(\mathbb{F}_5)\) is \(1 + 1 + 12 + 12 + 12 + 12 + 20 + 20 + 30\).

*This solution is based on hints provided by a contributor who prefers to remain anonymous.*We will assume that \(p\) is an odd prime; the special case \(p = 2\) can be dealt with separately. The order of \(SL_2(\mathbb{F}_p)\) is \(\frac{1}{p-1}(p^2 - 1)(p^2 - p) = p(p+1)(p-1)\).

As we saw in Problem 7.M.12, the number of solutions to the equation \(x^2 + axy + y^2 = 1\) in \(\mathbb{F}_p\) is equal to the order of the centralizer of the matrix \[ M_a = \begin{pmatrix} 0 & -1 \\ 1 & a \end{pmatrix} \] as ordered pairs \((x, y)\) satisfying \(x^2 + axy + y^2 = 1\) (then called \((b, d)\)) are in one-to-one correspondence with solutions to one-to-one correspondence with solutions to \eqref{eqn:M_a_comm}. To find \(|Z(M_a)|\), we note that the characteristic polynomial of \(M_a\) is \(\lambda^2 - a\lambda + 1\) and analyze the following cases into which \(M_a\) might fall:

Case 1: \(M_a\) has two distinct eigenvalues in \(\mathbb{F}_p\). Since \(\det M_a = 1\), these two eigenvalues must be \(\lambda\) and \(\lambda^{-1}\) for some \(\lambda \in \{2, \ldots, p - 2\}\). In this case, \(M_a\) is similar to the matrix \(\diag(\lambda, \lambda^{-1})\), so \(|Z(M_a)| = |Z(\diag(\lambda, \lambda^{-1}))|\). But the latter matrix commutes only with diagonal matrices, so its centralizer in \(SL_2(\mathbb{F}_p)\) has order \(p - 1\).

This case occurs when the discriminant of the characteristic polynomial, namely \(a^2 - 4\), is a nonzero perfect square in \(\mathbb{F}_p\). Although it's difficult to characterize the values of \(a\) for which this happens, for the purposes of part (d), we will only be interested in the total number of such \(M_a\) that fall into Case 1. Indeed, when we rewrite the characteristic polynomial in the form \(\lambda(\lambda - a) + 1\), we see that each \(\lambda \in \{2, \ldots, p-2\}\) has exactly one value of \(a\) for which the characteristic polynomial vanishes, giving a total of \((p-3)/2\) values of \(a\) for which \(M_a\) falls into Case 1, since each such \(a\) is shared by two values of \(\lambda\).

Case 2: \(M_a\) has a single eigenvalue in \(\mathbb{F}_p\) and is diagonalizable. Then \(M_a\) is a diagonal matrix, which is impossible.

Case 3: \(M_a\) has a single eigenvalue in \(\mathbb{F}_p\) and is defective. Again, this single eigenvalue must be 1 or -1. Suppose it is 1. Then \(N_1\) is the Jordan form of \(M_a\), so by Lemma 2, \(|Z(M_a)| = |Z(N_1)| = 2p\). If the single eigenvalue is -1, then the Jordan form of \(M_a\) is \(-N_{-1}\), and a similar result holds.

This case occurs when the discriminant vanishes, that is, \(a^2 - 4 = 0\), which occurs precisely when \(a = \pm 2\). When \(a = +2\) we get the double eigenvalue \(-1\), while when \(a = -2\) we get the double eigenvalue \(+1\).

Case 4: \(M_a\) has no eigenvalues in \(\mathbb{F}_p\), that is, there is no \(\lambda \in \mathbb{F}_p\) for which \(\lambda^2 - a\lambda + 1 = 0\), because \(a^2 - 4\) is not a perfect square in \(\mathbb{F}_p\). The number of values of \(a\) for which this occurs is obtained by subtracting the counts from Case 1 and Case 3 from \(p\), giving \((p-1)/2\).

For this case, in order to count the solutions to \(x^2 + axy + y^2 = 1\), we use Artin's hint, setting \(y = kx + 1\), which will be possible whenever \(x \ne 0\). The equation then becomes \((k^2 + ak + 1)x^2 + (a + 2k)x = 0\), which potentially has a nonzero solution \(x = -\frac{a + 2k}{k^2 + ak + 1}\). Indeed, the denominator cannot vanish for any value of \(k\), for if it did, then \(\lambda = -k\) would be a solution to the equation \(\lambda^2 - a\lambda + 1 = 0\), which by assumption does not occur. Thus, each value of \(k\) other than \(-a/2\) yields a nonzero value of \(x\) and a corresponding value of \(y\) solving \(x^2 + axy + y^2 = 1\), for a total of \(p - 1\) such solutions. When \(x = 0\), there are obviously exactly two solutions, namely \((0, \pm 1)\). Thus, the total number of solutions is \(p + 1\).

Case 1 above yields \((p-3)/2\) conjugacy classes, each of order \(|SL_2(\mathbb{F}_p)|/(p-1) = p(p+1)\). Case 3 yields 2 further conjugacy classes of order \(|SL_2(\mathbb{F}_p)|/(2p) = (p+1)(p-1)/2\). Case 4 yields \((p-1)/2\) further conjugacy classes of order \(|SL_2(\mathbb{F}_p)|/(p+1) = p(p-1)\). These \(p\) conjugacy classes are all disjoint since no two \(M_a\)'s are similar to each other, as they have unique traces. They therefore account for a total of \(p(p+1)(p-3)/2 + (p+1)(p-1) + p(p-1)^2/2 = p^3 - p^2 - p - 1\) elements. There are also two singleton conjugacy classes \(\{+I\}\) and \(\{-I\}\). This leaves \(p(p+1)(p-1) - (p^3 - p^2 - p + 1) = p^2 - 1\) elements unaccounted for.

Recalling part (b), let \(b\) be some element of \(\mathbb{F}_p\) that is not a quadratic residue; it's easy to see that such an element always exists. By Lemma 1, the matrices \(N_1\) and \(N_b\) are not conjugate in \(SL_2(\mathbb{F}_p)\). Both have trace 2, so at most one of them can be conjugate to one of the \(M_a\)'s, namely \(M_2\). Again, we won't need to determine which it is; Lemma 2 implies that the other, also having a centralizer of order \(2p\), belongs to a separate conjugacy class of size \((p+1)(p-1)/2\). Finally, \(-N_1\) and \(-N_b\) are likewise not conjugate to each other but each has a centralizer of order \(2p\) and trace \(-2\), so at least one of them is not conjugate to \(M_{-2}\) and belongs to a separate conjugacy class of size \((p+1)(p-1)/2\).

This accounts for all elements; the class equation of \(SL_2(\mathbb{F}_p)\) is: \[ p(p+1)(p-1) = 1 + 1 + 4 \times \frac{(p+1)(p-1)}{2} + \frac{p-1}{2} \times p(p-1) + \frac{p-3}{2} \times p(p+1) \] where the notation \(m \times n\) indicates \(m\) conjugacy classes of size \(n\).