Mathematics 403 Exam solutions

1. Given the figure, identify the point group.

Let p be the center of one of the ``empty'' hexagons. Then you can rotate about p by the angle $2\pi/6$ and by any of its multiples. The figure has no reflections or glide reflections for symmetries, so the point group is isomorphic to

$$C_{6} \cong \{ 1, \rho_{2\pi/6}, \rho_{4\pi/6}, \rho_{6\pi/6}, \rho_{2\pi/6}, \rho_{8\pi/6}, \rho_{10\pi/6}\}.$$

(Notice that since the translation group is a lattice, then the point group can't be C₅.)

2. What are the possible point groups when the translation group is a lattice?

C_n or D_n, with n=1, 2, 3, 4, or 6.

3. Let G be a finite group of rotations of the plane about the origin. Show that G is cyclic.

Let $\theta$ be the smallest positive angle of rotation in G. I claim that every element of G is a power of $\rho_{\theta}$; equivalently, I claim that if $\rho_{\alpha}$ is in G, then $\alpha$is a multiple of $\theta$. If $\alpha$ is not a multiple of $\theta$, then $\alpha$ lies strictly between two adjacent multiples of $\theta$: $k \theta < \alpha < (k+1) \theta$. Therefore, $0 < \alpha - k \theta < \theta$. Since $\rho_{\alpha}$ and $\rho_{\theta}$ are both in G, then so is

$$\rho_{\alpha} \rho_{\theta}^{-k} = \rho_{\alpha - k\theta}.$$

But this is a rotation in G by a smaller angle than $\theta$. This is a contradiction, so the assumption that $\alpha$ is not a multiple of $\theta$ must be wrong.

4. Identify the group $\Aut (C_{8})$.

Let $C_{8} = \{1, x, x^{2}, \ldots, x^{7} \}$. Every homomorphism $\phi$ from C₈ to itself is determined by where x goes: if $\phi(x) = x^{k}$, then $\phi(x^{2}) = (x^{k})^{2} = x^{2k}$, and more generally, $\phi(x^{i}) = x^{ik}$. So there are eight different homomorphisms from C₈ to itself; such a homomorphism is an automorphism if it sends x to an element of order 8. There are four elements of order 8--x, x³, x⁵, and x⁷--so there are four automorphisms:

• e, defined by e(xⁱ) = xⁱ.
• $\alpha$, defined by $\alpha(x^{i}) = x^{3i}$.
• $\beta$, defined by $\beta(x^{i}) = x^{5i}$.
• $\gamma$, defined by $\gamma(x^{i}) = x^{7i}$.

Thus $\Aut (C_{8}) = \{e, \alpha, \beta, \gamma\}$. To finish the problem, I need to specify the group structure on this set. For example, to compute $\alpha^{2} = \alpha \circ \alpha$, I look at

$$(\alpha \circ \alpha) (x) = \alpha (\alpha (x)) = \alpha (x^{3}) = x^{9} = x.$$

Since $\alpha \circ \alpha$ sends x to itself, then $\alpha \circ \alpha = e$. Similarly, $(\beta \circ \beta) (x) = x^{25} = x$, so $\beta \circ \beta = e$; and $(\gamma \circ \gamma) (x) = x^{49} = x$, so $\gamma \circ \gamma = e$. Since this is a group of order 4 in which no element has order 4, it must be isomorphic to the Klein 4 group, $C_{2} \times C_{2}$.

5. What are the orbits for the action of O(2) on the plane R²?

If A is orthogonal, then multiplication by A is a rigid motion that fixes the origin: it is either a rotation about the origin or a reflection across a line through the origin. Hence (as we saw in Section 4.5), orthogonal matrices preserve length: the length of Av is the same as the length of v; thus every vector in the orbit of v has the same length as v. Furthermore, if v and w are two vectors with the same length, there is a rotation that carries v to w; hence they are in the same orbit.

So the orbit of any vector v is

$$O_{v} = \{ w : \vert w\vert = \vert v\vert \}.$$

In other words, the orbits are the concentric circles around the origin.

6(a). What is the class equation of C₆?

Since $C_{6} = \{1, x, x^{2}, x^{3}, x^{4}, x^{5} \}$ is abelian, every element is conjugate only to itself: x^j xⁱ x^-j = xⁱ. Thus the conjugacy class of xⁱ just contains xⁱ, so the decomposition of the group into conjugacy classes is

$$C_{6} = \{1\} \cup \{x\} \cup \{x^{2}\} \cup \{x^{3}\} \cup \{x^{4}\} \cup \{x^{5}\},$$

and the class equation is

6 = 1 + 1 + 1 + 1 + 1 + 1.

6(b). What is the class equation of D₅?

Let's start with the conjugacy class of y; this is the set of all things of the form gyg^-1, where g is an element of D₅. Let g=1; then 1y1^-1 = y, so y is conjugate to itself. (This is true in any group: the conjugacy class of any element a always contains a.) I could let g=x, then g=x², etc., but things will be fastest if I let g=xⁱ:

xⁱ y x^-i = xⁱ xⁱ y = x²ⁱ y.

So I get these elements, as i goes from 1 to 4: x²y, x⁴y, x⁶y = xy, x^8y = x³y. So I know that the conjugacy class of y contains at least these elements: $\{y, xy, x^{2} y, x^{3} y, x^{4} y\}$. I claim that it doesn't contain anything else. Here are two different reasons: you can either compute (xⁱy) y (xⁱy)^-1 and see that you get the same things, or you can observe that the centralizer of y is $Z(y) = \{1,y\}$; since the order of the centralizer times the order of the conjugacy class is the order of the whole group, then the conjugacy class contains exactly 5 elements.

On to the conjugacy class of x: xⁱ x x^-i = x, and $(x^{i}y) x (x^{i} y)^{-1} = x^{i}y x y x^{-i} = \ldots = x^{-1} = x^{4}$. So $C_{x} = \{x, x^{4} \}$.

Similarly, $C_{x^{2}} = \{x^{2}, x^{3}\}$.

As always, $C_{1} = \{1\}$.

Thus

$$D_{5} = \{1\} \cup \{x, x^{4}\} \cup \{x^{2}, x^{3}\} \cup \{y, xy, x^{2} y, x^{3} y, x^{4} y\},$$

and the class equation is

10 = 1 + 2 + 2 + 5.

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