Brian Bi
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Section 7.2. The Class Equation

Let \(G\) be a finite nonabelian group and let \(C\) be one of its conjugacy classes. Then \(|G|/|Z(G)| = k|C|\) for some integer \(k > 1\).

If \(|C| = 1\) then this is immediate. Otherwise, let \(x \in C\); it is clear that \(x \notin Z(G)\). Then \(|C| = |G|/|Z(x)|\). But \(Z(x)\) contains at least one element that is not in \(Z(G)\), namely \(x\) itself. By Lagrange's theorem, \(|Z(x)| = k|Z(G)|\) for some integer \(k > 1\). The desired result follows.

Exercise 7.2.1

  1. Write \[ P = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \qquad A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \] Then \begin{align*} PA &= \begin{pmatrix} a & a+b \\ c & c + d \end{pmatrix} \\ AP &= \begin{pmatrix} a+c & b+d \\ c & d \end{pmatrix} \end{align*} therefore \(AP = PA\) iff \(c = 0, a = d\). Therefore \(\det P = a^2\), and there are two choices for \(a\) and three for \(b\). The centralizer of \(A\) therefore has order 6. The order of \(GL_2(\mathbb{F}_3)\) is \((3^2 - 1)(3^2 - 3) = 48\), so the conjugacy class of \(A\) has order 8.
  2. We previously found (Exercise 6.7.10(b)) that the centralizer has order 16 and the orbit has size 20.

Exercise 7.2.2 The centralizer of \(x\) has order \(21 \div 3 = 7\). Now the subgroup generated by \(x\) is a subgroup of the centralizer, so the former must have order 1 or 7. It can't have order 1 because then \(x\) would be the identity, whose conjugacy class contains only itself. Therefore \(x\) has order 7.

Exercise 7.2.3 Let \(x\) be a member of the conjugacy class of order 4. Then the centralizer \(Z(x)\) has order 3. The centre \(Z(G)\) is a subgroup of \(Z(x)\), so the former must have order 1 or 3. Suppose \(Z(G)\) has order 3; then \(Z(G) = Z(x)\). But \(Z(x)\) must also have \(\langle x \rangle\) as a subgroup, which must also have order 3, so \(Z(G) = \langle x \rangle\). If this were the case, then \(x\) would have a conjugacy class of size 1, which is a contradiction. Thus, \(Z(G)\) can only be trivial.

Exercise 7.2.4 If \(x = gyg^{-1}\), then \(x^n = (gyg^{-1})^n = gy^ng^{-1}\). That is, two conjugate elements of \(G\) are mapped by \(\varphi\) to two conjugate elements of \(G\), or, put another way, the image of each conjugacy class of \(G\) under \(\varphi\) is contained within a single conjugacy class. Moreover, for every \(x \in G\), the mapping \(\varphi: C(x) \to C(x^n)\) is surjective, since every \(gx^ng^{-1} \in C(x^n)\) occurs as \((gxg^{-1})^n\). It is not necessarily injective, however; for example, for \(n = |G|\), conjugacy classes of all sizes will be mapped onto \(\{1\}\).

Exercise 7.2.7 The class equation \(1 + 1 + 1 + 2 + 5\) describes a group whose centre has order 3, which doesn't divide 10, so this is impossible. The class equation \(1 + 2 + 3 + 4\) is also impossible because a group of order 10 can't have a conjugacy class of order 3, again because 3 doesn't divide 10. The class equation \(1 + 1 + 2 + 2 + 2 + 2\) is ruled out by the Lemma. Finally, the class equation \(1 + 2 + 2 + 5\) is actually valid; it is the class equation of \(D_5\). (This claim can be verified by straightforward computation; see Exercise 7.2.9(c).)

Exercise 7.2.8

  1. The class equation for a group of order 8 can only contain the numbers 1, 2, 4, and 8. It must contain 1, so it can't contain 8. The number of 1's must divide 8. The class equation \(1 + 1 + 1 + 1 + 1 + 1 + 1 + 1\) describes an abelian group. If there are four ones, then the class equation must be \(1 + 1 + 1 + 1 + 2 + 2\) or \(1 + 1 + 1 + 1 + 4\). Both of these can are ruled out by the Lemma. The class equation can't contain only one one (Proposition 7.3.1). So it must contain two ones and be either \(1 + 1 + 2 + 2 + 2\) or \(1 + 1 + 2 + 4\). The latter is ruled out by the Lemma. Thus, a nonabelian group of order 8 can only have class equation \(1 + 1 + 2 + 2 + 2\). As long as there exists a nonabelian group of order 8, its class equation must be \(1 + 1 + 2 + 2 + 2\). And we know of two examples of such groups, namely \(D_4\) and the quaternion group.
  2. The class equation can only contain the numbers 1, 3, and 7, and not all the numbers can be 1 for a nonabelian group. If the centre has order 7, then by inspection, the only possible class equation is \(1 + 1 + 1 + 1 + 1 + 1 + 1 + 7 + 7\), but this is ruled out by the Lemma. Now consider if the centre has order 3; again, by inspection, the only possible class equation is \(1 + 1 + 1 + 3 + 3 + 3 + 3 + 3 + 3\). This is also ruled out by the Lemma. The only remaining possibility that we haven't ruled out is that the centre is trivial, so the remaining numbers sum to 20. By inspection, the only possible class equation is then \(1 + 3 + 3 + 7 + 7\); if any nonabelian group of order 21 exists, this must be its class equation. It isn't obvious that such a group exists; it can be found by trial and error, or constructed as the semidirect product \(C_7 \rtimes C_3\), which hasn't been covered yet at this point in the text.

Exercise 7.2.9

  1. We argued in Exercise 7.2.8(a) that a nonabelian group of order 8 must have the class equation \(1 + 1 + 2 + 2 + 2\).
  2. Ditto.
  3. The five reflections in \(D_5\) are all conjugate to each other, since \(\rho^k r \rho^{-k} = \rho^k \rho^k r = \rho^{2k} r\) and the five choices of \(k\) each give a different reflection \(\rho^{2k}r\). A rotation can't be conjugate to a reflection, because the conjugate of an orientation-preserving element of \(D_5\) must be orientation-preserving (if the conjugation is by an orientation-reversing element, both that element and its inverse will reverse orientation, so these two reversals will cancel). The identity element belongs to its own conjugacy class. We claim that \(\rho\) and \(\rho^4\) form a conjugacy class. To see this, observe that \(r \rho r^{-1} = \rho^{-1} r r = \rho^{-1}\) and likewise \(r \rho^{-1} r^{-1} = \rho\). The rotations all commute with each other, so the conjugation of \(\rho\) by another rotation is just \(\rho\). And \(\rho, r\) generate \(D_5\), so if conjugation by these two elements fixes the subset \(\{\rho, \rho^4\}\), then conjugation by the entire group \(D_5\) does also. A similar argument establishes that \(\{\rho^2, \rho^3\}\) is a conjugacy class. In conclusion, the class equation is \(1 + 2 + 2 + 5\).
  4. Let \(G\) be the group so described. The order of \(G\) is 12, since we can choose any nonzero value for the top-left entry, any nonzero value for the bottom-right entry, and any value for the top-right entry. It is known that the diagonal elements of the product of two upper triangular matrices are just the products of the corresponding diagonal entries in the two factors. This implies that conjugate elements of \(G\) match along the diagonal. Divide \(G\) up into 4 subsets \(\{G_{ab} \mid a, b \in \{1, 2\}\}\) where \(G_{ab}\) consists of the three matrices with top-left entry \(a\) and bottom-right entry \(b\). Each \(G_{ab}\) must be a union of conjugacy classes. We now compute the centre of \(G\). A quick calculation shows that the two matrices \(\begin{pmatrix} a & b \\ 0 & c \end{pmatrix}\) and \(\begin{pmatrix} d & e \\ 0 & f \end{pmatrix}\) commute iff \(b(d-f) = e(a-c)\). This implies that if \(b = a - c = 0\), then the first matrix is central. If either \(b\) or \(a - c\) is nonzero, then we can always choose some \(e, d-f\) that don't satisfy this equation, and we can always construct some matrix in \(G\) with those \(e\) and \(d-f\) values. The centre of \(G\) therefore consists of the two diagonal matrices \(I, 2I\). Since the two elements of \(G_{11}\) other than \(I\) aren't central, they must be contained in conjugacy classes of order at least 2, which implies that those two elements form a single conjugacy class. The same holds for the two elements of \(G_{22}\) other than \(2I\). The three elements of \(G_{12}\) must form a single conjugacy class since there is no other way to partition a set of 3 elements without using any singleton; likewise the three elements of \(G_{21}\) must form a single conjugacy class. In conclusion, the class equation of \(G\) is \(1 + 1 + 2 + 2 + 3 + 3\).

Exercise 7.2.12 Suppose \(G\) has only one conjugacy class. The identity element always forms its own conjugacy class, so \(G\) must be the trivial group. Now suppose \(G\) has exactly two conjugacy classes. Then the class equation of \(G\) must be \(1 + x\), where \(x\) divides the order of the group \(1 + x\). This can only occur when \(x = 1\), so \(G \cong C_2\). Finally, suppose \(G\) has exactly three conjugacy classes. Write the class equation as \(1 + x + y\), with \(x \le y\). If \(x = 1\), then the order of the group is \(2 + y\), which must be a multiple of \(y\), therefore \(y\) can only be 1 or 2. The class equation \(1 + 1 + 1\) is that of \(C_3\), whereas the class equation \(1 + 1 + 2\) is impossible by an argument used in Exercise 7.2.8. Now consider \(x > 1\). We must have \(y \divides 1 + x + y\), so \(1 + x + y\) is at least \(2y\). If \(1 + x + y \ge 3y\), then this contradicts \(x \le y\). So \(1 + x + y = 2y\), or \(y = 1 + x\). The order of the group is then \(2 + 2x\). If \(x \divides 2 + 2x\) and \(x > 1\), then \(x\) can only be 2, corresponding to the class equation \(1 + 2 + 3\). We previously classified groups of order 6 and found that the only nonabelian group of order 6 is \(S_3\). We conclude that the only finite groups with at most three conjugacy classes are \(C_1, C_2, C_3, S_3\).

Exercise 7.2.13 A normal subgroup is a union of conjugacy classes. The group \(G\) has odd order so its class equation can't contain the numbers 2 or 4. \(N\) is therefore a union of conjugacy classes that have sizes 1, 3, or 5. One of these must have size 1, so none can have size 5. The only two ways to write 5 as a sum of 1's and 3's are 1, 1, 1, 1, 1 and 1, 1, 3. We claim that the case 1, 1, 3 is impossible. If \(N\) were a union of conjugacy classes of sizes 1, 1, 3, then let \(n\) be the element that belongs to the singleton that doesn't contain the identity. That is, \(n\) is central in \(G\). Since \(N\) has order 5, it must be cyclic, so the remaining three elements are \(n^2, n^3, n^4\). But these would also be central in \(G\), which contradicts them forming a conjugacy class of order 3. From this we see that the only possibility is 1, 1, 1, 1, 1: each element of \(N\) belongs to its own conjugacy class in \(G\), that is, \(N\) is contained in the centre of \(G\).

Exercise 7.2.14

  1. Yes, since \(G\) has order 20 and has a conjugacy class of order 4, it must have some centralizer subgroup with order 5. Now, a subgroup of order 5 is necessarily the subgroup generated by some element of order 5 in \(G\). If \(x \in G\) belongs to a conjugacy class of size 5, then its centralizer has order 4 and contains the subgroup generated by \(x\), therefore \(x\) can't have order 5. The only elements of \(G\) that can possibly have order 5 are therefore the four elements of the single conjugacy class of size 4. We know there are at least four elements of \(G\) with order 5 since we already know \(G\) has a subgroup of order 5; therefore the four elements in the conjugacy class of size 4 are precisely the set of elements of order 4 in \(G\). This implies that there is exactly one subgroup of \(G\) with order 5, since any such subgroup has to contain 4 elements of \(G\) that have order 4. Since there is only one subgroup of \(G\) with order 5, any conjugate to this subgroup has to be that subgroup itself. Therefore the subgroup of order 5 is normal.
  2. Yes, since \(G\) has order 20 and has a conjugacy class of order 5, it must have some centralizer subgroup with order 4. Now a normal subgroup of \(G\) would consist of a union of conjugacy classes, one of which has size 1, so its order would be the sum of some terms in \(G\)'s class equation including 1. This can't be 4, so \(G\) has no normal subgroup of order 4.

Exercise 7.2.16 For every \(g \in G\), \(\varphi(gxg^{-1}) = \varphi(g)\varphi(x)\varphi(g)^{-1}\) lies in the conjugacy class of \(\varphi(x)\). Moreover, every element of \(C'\) takes the form \(g'\varphi(x)g'^{-1}\) where \(g' \in G'\). For such an element, let \(g \in G\) be such that \(\varphi(g) = g'\); we know such \(g\) exists because \(\varphi\) is surjective. Then \(\varphi(gxg^{-1}) = g'\varphi(x)g'^{-1}\). But \(gxg^{-1} \in C\). Therefore \(\varphi\) maps \(C\) surjectively to \(C'\).

We are also asked to show that \(|C'|\) divides \(|C|\). I will assume that \(G\) is finite. (I am not entirely sure whether the claim is even true in the case where \(G\) is infinite but \(C\) is finite.) Then \begin{align*} \frac{|C|}{|C'|} &= \frac{|G|/|Z(x)|}{|G'|/|Z(\varphi(x))|} \\ &= \frac{|G|/|G'|}{|Z(x)|/|\varphi(Z(x))|} \frac{|Z(\varphi(x))|}{|\varphi(Z(x))|} \\ &= \frac{|\ker \varphi|}{|Z(x) \cap \ker \varphi|} \frac{|Z(\varphi(x))|}{|\varphi(Z(x))|} \end{align*} Now \(Z(x) \cap \ker \varphi\) is a subgroup of \(\ker\varphi\) so the first factor in this product is an integer; and \(\varphi(Z(x))\) is a subgroup of \(Z(\varphi(x))\) so the second factor in the product is an integer. Therefore \(|C|/|C'|\) is an integer.

Exercise 7.2.17 Let \(G\) be a group of order \(pq\). Consider the following cases:

  1. \(p = q\). That is, \(|G| = p^2\). The only possible orders of non-identity elements in \(G\) are \(p\) and \(p^2\). But if an element has order \(p^2\) then its \(p\)th power has order \(p\), so we are done.
  2. \(p \neq q\) and there is a conjugacy class of size \(q\) in \(G\). Let \(x\) belong to such a conjugacy class. Then \(|Z(x)| = p\). This implies that \(Z(x)\) is a cyclic group and its generators have order \(p\).
  3. \(p \neq q\), there is no conjugacy class of size \(q\) in \(G\), and \(G\) is nonabelian. The class equation consists of \(1\)'s and \(p\)'s. The number of \(1\)'s in the class equation must then be a multiple of \(p\), so \(|Z(G)|\) can only be \(p\). Then, each generator of \(Z(G)\) has order \(p\).
  4. \(p \neq q\) and \(G\) is abelian. Pick \(x \in G \setminus \{1\}\). If \(x\) has order \(p\) then we're done. If \(x\) has order \(pq\) then \(x^q\) has order \(p\). Otherwise, \(x\) has order \(q\). Let \(H = \langle x \rangle\). Pick \(y \in G \setminus H\). Again, if \(y\) has order \(pq\) then we're done. Otherwise \(y\) has prime order. Let \(K = \langle y \rangle\). By Proposition 2.11.4(c), \(HK\) is a subgroup of \(G\), which itself contains both \(H\) and \(K\) as proper subgroups; by Lagrange's theorem, the only possible order of \(HK\) is \(pq\), so \(G = HK\). Also, \(H \cap K\) is a subgroup of both \(H\) and \(K\) so its order can only be 1 (or \(q\) in the case where \(|K| = q\), but that's not possible since \(H \neq K\)). By Proposition 2.11.4(d), \(G\) is isomorphic to \(H \times K\). Therefore \(y\) has order \(p\).

Exercise 7.2.18

  1. They are conjugate for all \(d\), by the element \(\diag(1, -1)\).
  2. This is just an algebra problem. Write \begin{equation*} \begin{pmatrix} e & f \\ g & h \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & d \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & d \end{pmatrix} \begin{pmatrix} e & f \\ g & h \end{pmatrix} \end{equation*} and deduce that this holds iff \(f = g\) and \(h = -(e + df)\). Since every pair \(e, f \in \mathbb{R}\) uniquely determines \(g, h\), we can rewrite the condition for this matrix to have determinant 1 in terms of \(e, f\) alone, giving \(e^2 + def + f^2 = -1\). This will have a real solution iff \(|d| > 2\).