Brian Bi
$\DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\char}{char} \DeclareMathOperator{\lcm}{lcm} \newcommand\divides{\mathbin |} \newcommand\ndivides{\mathbin \nmid} \newcommand\d{\mathrm{d}} \newcommand\p{\partial} \newcommand\C{\mathbb{C}} \newcommand\N{\mathbb{N}} \newcommand\Q{\mathbb{Q}} \newcommand\R{\mathbb{R}} \newcommand\Z{\mathbb{Z}} \newcommand\pref[1]{(\ref{#1})} \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\Gal}{Gal}$

## 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

1. 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 $$N \subseteq HN$$.
2. 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

1. 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 $$\overline{G}$$.
2. 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

1. $$p_1$$ 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 $$p_1 p_2 = p_2 p_1 = C_1 C_2 \ldots C_m D_1 D_2 \ldots D_n$$.
2. 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:

1. The conjugacy class consisting only of $$I$$;
2. The conjugacy class consisting only of $$-I$$;
3. The conjugacy class of elements of order 2 with determinant −1 that are of K-type;
4. The conjugacy class of elements of order 2 with determinant −1 that are not of K-type.

Problem 7.M.12

1. Let $$M_a = \begin{pmatrix} 0 & -1 \\ 1 & a\end{pmatrix}$$. Suppose $$M_a \begin{pmatrix} b & c \\ d & e \end{pmatrix} = \begin{pmatrix} b & c \\ d & e \end{pmatrix} M_a \label{eqn:M_a_comm}$$ 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.
• $$a = 1$$: $$(1, 4), (4, 1)$$
• $$a = 2$$: $$(1, 3), (2, 2), (2, 4), (3, 1), (3, 3), (4, 2)$$
• $$a = 3$$: $$(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3)$$
• $$a = 4$$: $$(1, 1), (4, 4)$$
So the centralizers have orders 4, 6, 10, 10, 6 in those respective cases. We now determine the subgroup structure.
• $$a = 0$$: 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 $$Z(M) = \langle -M \rangle \cong C_6$$.
2. 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 $$1 + 1 + 12 + 12 + 12 + 12 + 20 + 20 + 30$$.

3. 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 $$p + 1$$.

4. 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 $$n$$.