Brian Bi

## Section 7.4. The Class Equation of the Icosahedral Group

Exercise 7.4.1 In Theorem 7.4.4 it was proved that the image of the action of $$I$$ on the set of five cubes is $$A_5$$. The stabilizer of a single cube is therefore the subgroup of $$A_5$$ that leaves one element, say, index 1, fixed. If $$\sigma \in A_5$$ satisfies $$\sigma(1) = 1$$ then the restriction $$\sigma|_{\{2, 3, 4, 5\}}$$ must also be even, and each such even permutation can be uniquely extended to an even permutation on the entire set that fixes 1. Therefore the desired subgroup is simply $$A_4$$.

Exercise 7.4.2 Yes. In fact, we will prove the more general statement that for all $$n \geq 5$$, $$A_n$$ is the only proper normal subgroup of $$S_n$$. Let $$N$$ be a normal subgroup of $$S_n$$. Then the set $$N'$$ consisting only of the even permutations in $$N$$ would be a subgroup of $$N$$. It would also be normal since the conjugate of an even permutation is also another even permutation. Therefore $$N'$$ would be a normal subgroup of $$S_n$$ that is contained within the $$A_n$$ subgroup. Therefore $$N'$$ would be a normal subgroup of $$A_n$$. Since $$A_n$$ is simple, either $$N'$$ is trivial or $$N' = A_n$$.

If $$N'$$ is trivial, then either $$N$$ is trivial or else $$N$$ consists only of the identity and odd permutations. Let $$\sigma$$ be an odd permutation in $$N$$. Clearly $$\sigma^2$$ must be the identity. If there is any other odd permutation in $$N$$, then the product of these two odd permutations would be an even permutation that isn't the identity, which is a contradiction. So $$N = \{1, \sigma\}$$. It's easy to see that such $$N$$ could not be normal.

If $$N'$$ is $$A_n$$, then either $$N = A_n$$ or else $$N$$ contains some odd permutation $$\sigma$$, the set $$\sigma A_n$$ would consist of the $$n!/2$$ distinct odd permutations of $$S_n$$, and therefore $$N$$ would be all of $$S_n$$.

We conclude that the only proper normal subgroup of $$S_n$$ is $$A_n$$.

Exercise 7.4.3 It was determined in the text that the elements of $$I$$ of order 2 belong to a conjugacy class of size 15 and each represents a 180 degree rotation around the axis joining two opposite edges. Let $$g$$ be such an element associated to edge $$e$$. The centralizer $$Z(g)$$ must have order 4 and since $$I$$ contains no elements of order 4, $$Z(g)$$ can only contain elements of order 2, implying that $$Z(g) \cong V_4$$. To actually describe the elements of this $$V_4$$ subgroup of $$I$$, observe that there are four edges perpendicular to $$e$$, which occur in two pairs of opposite edges that are also perpendicular to each other. They therefore have two associated rotations $$h, i$$ and the three rotations $$g, h, i$$ are 180 degree rotations around mutually perpendicular axes. Therefore $$g$$ commutes with $$h$$ and $$i$$ and the elements $$\{1, g, h, i\}$$ constitute the centralizer of $$e$$. This group is not the cyclic group of order 4 so it can only be the Klein four-group.

Exercise 7.4.4

1. Using Corollary 5.1.28(b) as in the similar calculation for the icosahedral group done in the text, if $$V$$ is a vertex of the tetrahedron then a rotation by $$2\pi/3$$ counterclockwise around the axis joining a vertex to the centroid is conjugate to each other such rotation. However in order to get the rotation by $$2\pi/3$$ clockwise, which is a rotation by $$2\pi/3$$ counterclockwise where the pole $$P$$ is above the opposite face, we would have to conjugate by a rotation that maps $$V$$ to $$P$$, which is not therefore in $$T$$. Therefore the four 120-degree clockwise rotations fall into a different conjugacy class from the four 120-degree counterclockwise rotations. The three 180 degree rotations are also seen to form a single conjugacy class of size 3. So the class equation is $$12 = 1 + 3 + 4 + 4$$.
2. Any subgroup of $$T$$ with 6 elements would have to be normal, but 6 is not the sum of any set of terms in the class equation including 1. So there is no subgroup of order 6. There is only one subgroup of order 4: such a subgroup cannot contain any of the elements of order 3, so it must consist of the identity and the three 180-degree rotations. Since this is the only subgroup of order 4 in $$T$$, it is normal.

Exercise 7.4.5

1. Using Corollary 5.1.28(b) as in the similar calculation for the icosahedral group done in the text, all six 90-degree rotations form a single conjugacy class, as do the eight 120-degree rotations. The six 180-degree rotations that have an axis passing through the midpoints of opposite edges of a cube are likewise conjugate to each other and the three 180-degree rotations that have an axis passing through the centres of opposite faces are conjugate to each other. These last two groups are separate conjugacy classes because by Corollary 5.1.28(b), if conjugation by a rotation $$B$$ mapped a rotation with an edge pole to a rotation with a face pole, then $$B$$ itself would have to map the edge pole to the face pole, so $$B \notin O$$. We conclude that the class equation is $$24 = 1 + 3 + 6 + 6 + 8$$.
2. There are two sets of terms in the class equation including the 1 that sum to a proper divisor of 24: namely, $$1 + 3$$ and $$1 + 3 + 8$$. From the analysis in part (a) we know that the conjugacy class of order 3 consists of the three rotations $$\rho_{f, \pi}$$ for $$f$$ a face and the conjugacy class of order 8 consists of the rotations $$\rho_{v, 2\pi/3}$$ for $$v$$ a vertex. If these sets $$G_4$$ and $$G_{12}$$ are subgroups of $$O$$, then they are normal. It is clear enough that $$G_4$$ is a group isomorphic to $$V_4$$; it's just the group consisting of the identity and three 180 degree rotations around mutually perpendicular axes. Meanwhile, the quickest way to characterize $$G_{12}$$ is to observe that each 120-degree rotation induces a 3-cycle on the other three body diagonals while each 180-degree rotation in $$G_{12}$$ exchanges two pairs of body diagonals, so each element of $$G_{12}$$ induces an even permutation of the four body diagonals. This implies that $$G_{12}$$ is isomorphic to the $$A_4$$ subgroup of $$O$$ when regarded as $$S_4$$. See Exercise 7.4.6(a).

Exercise 7.4.6

1. In Exercise 6.9.4 we showed that $$T \cong A_4$$. The proof that $$O \cong S_4$$ is even easier. Let $$S$$ be the set consisting of the four body diagonals of a cube. From considering the different kinds of rotations in $$O$$, it is clear that the only element of $$O$$ that fixes all four body diagonals is the identity element. That is, the action $$\varphi : O \to \operatorname{Perm}(S)$$ is faithful. But these two groups both have order 24, so $$O \cong S_4$$.
2. The 12 elements of $$O$$ that belong to the $$A_4$$ subgroup are also rotational symmetries of the inscribed tetrahedra; that is, the tetrahedral group arises as the $$A_4$$ subgroup of the octahedral group. The other 12 elements of $$O$$ exchange the two tetrahedra.

Exercise 7.4.7 Since $$|G| > |\operatorname{Perm}(S)|$$ where $$S$$ is the set that $$G$$ acts on, the permutation representation $$\varphi : G \to \operatorname{Perm}(S)$$ is not faithful. The kernel of $$\varphi$$ is therefore not trivial. It also isn't the entire group $$G$$ since we are told that $$G$$ acts nontrivially on $$S$$. Therefore $$\ker \varphi$$ is a proper normal subgroup of $$G$$.

Exercise 7.4.8

1. The centre is a subgroup of the centralizer, so it can only have order 1, 2, or 4.
2. By the Lemma in the solutions to Section 7.2, the order of the centre is $$|G|/(4k)$$ for some integer $$k > 1$$. Additionally, we can say that $$y \notin Z$$.

Exercise 7.4.9 We know $$x \in Z(x)$$. The order of $$x$$ must be $$p$$, $$q$$, or $$pq$$. If it is $$pq$$ then $$Z(x)$$ is just $$C_{pq}$$. Otherwise, without loss of generality, say the order of $$x$$ is $$p$$. Let $$H = \langle x \rangle$$ and take $$y \in Z(x) \setminus H$$. Let $$K = \langle y \rangle$$. Note that $$H$$ is a normal subgroup of $$Z(x)$$. By Proposition 2.11.4, $$HK$$ is a subgroup of $$Z(x)$$ which contains $$H$$ as a proper subgroup. By Lagrange's theorem, $$HK$$ must be all of $$Z(x)$$. But then $$Z(x)$$ is generated by $$x$$ and $$y$$, which commute with each other since $$y \in Z(x)$$; therefore $$Z(x)$$ is abelian.