Section 7.8. The Groups of Order 12
Exercise 7.8.1 \(S_3 \times C_2\) is nonabelian, so it's not isomorphic to \(C_4 \times C_3\) or to \(C_2 \times C_2 \times C_3\). It contains no elements of order 4, so it's not isomorphic to the fifth group in Theorem 7.8.1 (which is the dicyclic group \(\operatorname{Dic}_3\)). It contains 3 elements of order 3; \(A_4\) contains 8 such elements while \(D_6\) contains 3. So \(S_3 \times C_2 \cong D_6\).
Exercise 7.8.2
- We need to find \(x, y\) with \(x^4 = y^3 = 1, xyx^{-1} = y^2\). By trial and error, we can find an embedding into \(S_7\), with \(x = (1\ 2)(4\ 5\ 6\ 7), y = (1\ 2\ 3)\); since conjugation acts as relabelling cycles, \(xyx^{-1} = (2\ 1\ 3) = y^2\). Now we need to prove there is no embedding into \(S_n\) for \(n < 7\). First consider the case where \(y\) consists of a single 3-cycle; without loss of generality, let \(y = (1\ 2\ 3)\). In order for conjugation by \(x\) to reverse this cycle, it needs to map 123 to 132, 213, or 321; each of these is a transposition. Without loss of generality, let the transposition be \((1\ 2)\). So \(x\) must contain this transposition while fixing 3; if \(n \le 6\) then such \(x\) cannot have order 4, as there aren't enough elements left to form a 4-cycle. This rules out all \(n \le 5\). For \(n = 6\) there is an additional case to consider, namely that \(y\) contains two 3-cycles; without loss of generality, say \(y = (1\ 2\ 3)(4\ 5\ 6)\). The same reasoning as before shows that \(x\) can't transpose two elements of each 3-cycle while also having order 4. Another possibility is that \(x\) relabels the two cycles into each other's inverses. Without loss of generality, say \(x(1) = 4\). If \(x\) contains the tranposition \((1\ 4)\), then it must also swap 2 with 6 and 3 with 5, so would not have order 4. If \(x\) maps 1 to 4 in a 4-cycle, then it must map 4 to 2 or 3, respectively forcing \(x = (1\ 4\ 2\ 6\ 3\ 5)\) and \(x = (1\ 4\ 3\ 5\ 2\ 6)\), which both don't have order 4. So all \(n \le 6\) are ruled out.
(Thanks to Xingyou Song.) The most difficult part of this problem is finding a matrix \(y \in SL_2(\mathbb{F}_5)\) such that \(y^3 = I\). Diagonal matrices, and matrices similar to diagonal matrices, can be ruled out because the only cube root of 1 modulo 5 is 1 itself; an antidiagonal matrix such as the one in (7.8.3) wouldn't work either.
Here is a somewhat systematic approach. Hope that there exists \(y \in SL_2(\mathbb{F}_5)\) such that \(y^2 + y + I = 0\); if there is such \(y\), then it'll automatically be the case that \(y^3 - I = (y - I)(y^2 + y + I) = 0\). Complete the square to yield \((y + I/2)^2 + 3I/4 = 0\); in \(\mathbb{F}_5\) this simplifies to \((y + 3I)^2 = 3I\). Now look for square roots of \(3I\). We can take antidiagonal matrices as an ersatz: \[ \begin{pmatrix} 0 & a \\ b & 0 \end{pmatrix}^2 = \begin{pmatrix} ab & 0 \\ 0 & ab \end{pmatrix} \] So for example choosing \(a = 1, b = 3\) would work. Subtracting \(3I\) gives a solution for \(y\): \[ y = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} \] One can verify that \(y^3 = I\), and it appears that we have gotten lucky because \(\det y = 1\).
Now we can turn our attention to finding \(x\). Let's write down \(y^2\) explicitly: \[ y^2 = \begin{pmatrix} 2 & 4 \\ 2 & 2 \end{pmatrix} \] We might as well try a diagonal matrix for \(x\). The two entries must be inverses so that \(\det x = 1\), and \(\diag(1, 4)\) doesn't work because it has order 2, so instead let's try \(\diag(2, 3)\), which has order 4. Its inverse is \(\diag(3, 2)\). Indeed, \[ \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 2 & 4 \\ 2 & 2 \end{pmatrix} \] so we are done; the desired subgroup is generated by \[ x = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} \qquad y = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} \]
Exercise 7.8.3 Two of the groups of order 12, namely \(C_4 \times C_3\) and \(C_2 \times C_2 \times C_3\), are abelian, so the class equation is obvious. We previously determined that the class equation of \(A_4\) is \(12 = 1 + 3 + 4 + 4\). We now focus our attention on the remaining two groups.
The group \(D_6\) is generated by \(\rho, r\) with \(\rho^6 = r^2 = 1\) and \(r\rho = \rho^{-1}r\). This allows us to compute: \[ (\rho^i r^j)r(\rho^i r^j)^{-1} = \rho^i r^{j+1} r^{-j} \rho^{-i} = \rho^i r \rho^{-i} = \rho^i \rho^i r = \rho^{2i} r \] so the conjugacy class of \(r\) is \(\{r, \rho^2 r, \rho^4 r\}\). Likewise \begin{align*} (\rho^i)\rho r(\rho^{-i}) &= \rho^{i+1} \rho^i r = \rho^{2i+1}r \\ (\rho^i r)\rho r (\rho^i r)^{-1} &= \rho^i \rho^{-1} r r r \rho^{-i} = \rho^{i-1} r \rho^{-i} = \rho^{i-1} \rho^i r = \rho^{2i-1}r \end{align*} so the conjugacy class of \(\rho r\) is \(\{\rho r, \rho^3 r, \rho^5 r\}\). The rotation \(\rho^3\) is not conjugate to any of the reflections and is the only rotation of order 2, so it lies in its own conjugacy class. \(\rho\) and \(\rho^{-1}\) are conjugate to each other by \(r\), and are the only elements of order 6, so they are not conjugate to \(\rho^2\) and \(\rho^{-2}\), which have order 3 and are also conjugate to each other \(r\). Therefore the class equation is \(12 = 1 + 1 + 2 + 2 + 3 + 3\).
We turn lastly to the group \(\operatorname{Dic}_3\) which is generated by \(x\) and \(y\) with \(x^4 = y^3 = 1, xy = y^2 x\). We might start by observing that \(x^2\) commutes with \(y\), so \(x^2\) is central. Also, conjugation of \(y\) by some \(x^i y^j\) is equivalent to applying the conjugation by \(y^j\) first (yielding \(y\) again) and then the conjugation by \(x^i\), which will give either \(y\) or \(y^2\). So \(y\) and \(y^2\) form a conjugacy class. The conjugate of \(x\) by \(y\) is \(yxy^2 = yxyy = y(y^2 x)y = y y^2 y^2 x = y^2 x\) and likewise the conjugate by \(y^{-1}\) is \(yx\). The conjugation of \(x\) by some \(y^i x^j\) is equivalent to applying the conjugation by \(x^j\) first, yielding back \(x\), then the conjugation by \(y^i\), yielding \(x\), \(yx\), or \(y^2 x\). So \(\{x, yx, y^2 x\}\) form a conjugacy class. This implies that the inverses of these elements also form a single conjugacy class, namely \(\{x^3, yx^3, y^2 x^3\}\). Lastly, \(x(yx^2)x^3 = (y^2x)x^2 x^3 = y^2 x^2\), so the two elements \(yx^2, y^2 x^2\) form a conjugacy class. The class equation is therefore \(12 = 1 + 1 + 2 + 2 + 3 + 3\).
Exercise 7.8.4 For \(p = 2\) this is evident. Suppose \(|G| = 2p\) where \(p\) is a prime greater than 2. Then \(G\) contains exactly one Sylow \(p\)-subgroup, which is therefore normal. Call this subgroup \(H\). Since \(H\) has prime order, it is cyclic; let \(x\) be a generator of \(H\). Let \(y\) be some element of order 2. It is clear that \(G = \langle y \rangle H\), so knowing \(yxy^{-1}\) will be sufficient to determine the multiplication table. If \(yxy^{-1} = x\), then \(xy\) generates \(G\), that is, \(G \cong C_{2p}\). If \(yxy^{-1} = x^i\) for some \(i \not\equiv 1 \pmod{p}\), then we must have \(i^2 \equiv 1 \pmod{p}\). But \(\mathbb{Z}/p\mathbb{Z}\) is a field, so the equation \(i^2 - 1 = 0\) with \(i \in \mathbb{Z}/p\mathbb{Z}\) has at most two solutions, namely \(i = 1\) and \(i = -1\). So the other possibility is \(yx = x^{-1}y\). This defines the dihedral group \(D_p\).
Exercise 7.8.5
- By Sylow's third theorem, there is exactly one Sylow 7-subgroup, call it \(H\), which is therefore normal, and the number of Sylow 2-subgroups is either 1 or 7. Let \(K\) be a Sylow 2-subgroup. The intersection \(H \cap K\) must be trivial by Lagrange's theorem, and by Proposition 2.11.4, it is clear that \(HK = G\). If \(K\) is the only Sylow 2-subgroup, then it is normal, and \(G \cong H \times K\), making \(G\) abelian. We are given that \(G\) is nonabelian, therefore \(K\) is one of seven Sylow 2-subgroups.
- Write \(H = \langle x \rangle\) and \(K = \langle y \rangle\). Since \(H\) is normal, we must have \(yxy^{-1} = x^i\). But then \(i^4 \cong 1 \pmod{7}\), and there are only two solutions, namely \(i = \pm 1\). If \(i = 1\) then we would have \(G \cong C_4 \times C_7\), which we have already ruled out. Therefore \(i = -1\). This completely determines the multiplication table for the group.
Since \(yxy^{-1} = x^{-1}\), we have that \(y^2\) commutes with \(x\). Therefore \(x\) and \(y^2\) generate a subgroup \(I \subseteq G\) isomorphic to \(C_2 \times C_7 \cong C_{14}\), and that \(y\) doesn't belong to this subgroup (if it did, then \(y\) would have to commute with \(x\)). Let \(z\) be a generator of \(I\), for example, \(z = xy^2\). It is clear that \(G = \{1, y\}I\), and we can compute the conjugate, \(yzy^{-1} = y(xy^2)y^{-1} = x^{-1} y y^2 y^{-1} = x^{-1} y^2 = z^{-1}\). It's also clear that \(y^2 = z^7\), since 7 is the only element of \(C_{14}\) of order 2. These facts will make it easy to determine the orders of elements and conjugacy classes in \(G\).
The group \(I\) contains the identity, one element of order 2, namely \(z^7\), six elements of order 7, namely \(z^{2i}\) for \(i\) from 1 to 6, and six elements of order 14, namely the remaining powers of \(z\). There are seven Sylow 2-subgroups, and the 14 elements of order 4 contained among them must all be distinct, as each such element generates the corresponding Sylow 2-subgroup. Thus, in total there are: 1 element of order 1, 1 element of order 2, 6 elements of order 7, 6 elements of order 14, and 14 elements of order 4.
The element \(y\) has at least 7 conjugates (including itself) since its conjugates must generate the 7 Sylow 2-subgroups of \(G\). Since the centralizer of \(y\) contains \(K\), which is of order 4, the conjugacy class of \(y\) must have size at most 7. So the conjugacy class of \(y\) has size exactly 7. Each conjugate of \(y\) therefore belongs to a different Sylow 2-subgroup, so \(y\) is not conjugate to \(y^{-1}\). The element \(y^{-1}\) therefore belongs to a conjugacy class of size 7 that comprises the inverses of the elements of the conjugacy class of \(y\). The identity, of course, has its own conjugacy class, and so does \(z^7\), since it is the only element of order 2. All possible elements of \(G\) that don't commute with some \(z^j\) can be written in the form \(yz^i\). Conjugation of \(z^j\) by \(yz^i\) is equivalent to conjugating first by \(z^i\), which gives back \(z^j\), and then conjugating by \(y\), giving \(z^{-j}\). So for \(j \in \{1, 2, 3, 4, 5, 6\}\), there is a conjugacy class \(\{z^{\pm j}\}\). The class equation of \(G\) is therefore \(28 = 1 + 1 + 6 \cdot 2 + 7 + 7\).
Exercise 7.8.6
- There is exactly one Sylow 11-subgroup, which is therefore normal; call it \(H = \langle x \rangle\). Let \(K = \langle y \rangle\) be a Sylow 5-subgroup. Then \(H \cap K\) is trivial, and by Proposition 2.11.4, \(G = HK\). That is, \(x\) and \(y\) generate \(G\). The conjugate \(yxy^{-1}\) must lie in \(H\) since \(H\) is a normal subgroup; so \(yxy^{-1} = x^r\) for some \(r\) with \(1 \le r \le 10\).
- It must be that \(r^5 \cong 1 \pmod{11}\). By trying all \(r\) from 1 to 10, we find \(r \in \{1, 3, 4, 5, 9\}\).
- The choice of \(r = 1\) would yield an abelian group, which by Proposition 2.11.4 is isomorphic to \(C_5 \times C_{11}\). It turns out that all other choices are equivalent. If \(r = 4\) then \(y^4 x y^{-4} = x^{256} = x^3\), and \(y^4\) also has order 5, so we can replace \(y\) by \(y^4\). Likewise if \(r = 5\) then we can replace \(y\) with \(y^2\) and if \(r = 9\) then we can replace \(y\) with \(y^3\). So we can always pick \(y\) so that \(yxy^{-1} = x^3\). This completely determines the multiplication table of \(G\), so there are at most two isomorphism classes of groups of order 55, one of which is \(C_5 \times C_{11}\). If there really is a group of order 55 generated by \(x\) and \(y\) with \(x^{11} = y^5 = 1, yxy^{-1} = x^3\), then it is nonabelian, so it is not isomorphic to \(C_5 \times C_{11}\), so there are exactly two isomorphism classes. We can construct such a group as a subgroup of \(S_{11}\) by taking \(x = (1\ 2\ 3\ 4\ 5\ 6\ 7\ 8\ 9\ 10\ 11), y = (2\ 4\ 10\ 6\ 5) (3\ 7\ 8\ 11\ 9)\). It is obvious that \(x^{11} = y^5 = 1\) and we can compute the conjugate \(yxy^{-1}\) by relabelling; this turns out to be \((1\ 4\ 7\ 10\ 2\ 5\ 8\ 11\ 3\ 6\ 9)\), which is indeed \(x^3\). This establishes that \(x, y\) generate a subgroup of \(S_{11}\) of order at most 55, which is divisible by both 5 and 11; and is therefore exactly 55.
Note from Brian: This is not the first time we have had to construct a nonabelian group of order \(pq\); we previously did so in Exercises 7.7.6 and 7.7.9(c). We have done so by giving explicit subgroups of \(S_n\) isomorphic to the groups in question, but these examples motivate a much more general technique for constructing these groups. We have seen that in each case, \(G = HK\) so the elements of \(G\) are pairs consisting of an element of \(H\) and an element of \(K\), but the composition rule isn't the same one as in the direct product. Nevertheless, if we could give explicitly a composition rule for such pairs and show that it satisfied the group axioms, we wouldn't need to construct these subgroups of \(S_n\). With a commutation relation of the form \(yxy^{-1} = x^r\), the composition rule can be stated easily, but is cumbersome for performing algebraic manipulations. The more general technique is to observe that conjugation by each \(k \in K\) induces an automorphism \(\varphi_k\) on \(H\), and these automorphisms must be compatible with the composition rule of \(K\), that is, the composition rule can be defined in terms of some homomorphism \(\varphi : K \to \operatorname{Aut}(H)\). This construction is the outer semidirect product. This is, of course, well known, but there's only a finite amount of space in Artin's text, so it hasn't been covered at this point.