Brian Bi

## 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

1. 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.
2. (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

1. 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.
2. 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.
3. 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

1. 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 $$1 \le r \le 10$$.
2. It must be that $$r^5 \cong 1 \pmod{11}$$. By trying all $$r$$ from 1 to 10, we find $$r \in \{1, 3, 4, 5, 9\}$$.
3. 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.