Homework 9 M 373K
by Mark Lindberg (mal4549)
Recall that the point group of a subgroup G < Mn , call it G, is the image G := im(π|G ) of
G under the homomorphism π : Mn On (cf. Section 6.5). As we saw in class, we can analyze
G by considering both its point group and also its translational subgroup, L := ker(π|G ) ≤ G,
so that G/L ∼
= G. We caution that, while the translational subgroup is a subgroup, the point
group is not a subgroup in general; rather, it is a quotient (equivalently, a homomorphic
image).
In the first two problems, we analyze this in more detail and we will see, in computing
the symmetry of a plane figure, the reflections in the point group can mean either reflectional
or glide symmetry (so there need not be any reflections in the group of symmetries itself),
whereas the rotations in the point group must actually correspond to rotational symmetries
in the group itself.
1. Consider the group G generated by a single glide reflection. Show that the point group
is D1 = {id, r} ∼
= Z/2 for a single reflection r, whereas G ∼
= Z has no elements of finite
order. Deduce that in this case the point group cannot be realized as a subgroup of G.
Let the single glide reflection be tv r` , where v is parallel to `, and, indeed, choose an origin
on `, so that v lies on ` and is parallel to it. Then G = htv r`i, and all elements are of the
form (tv r`k . Consider the first few:
• k = 0: (tv r` )0 = id
• k = 1: tv r` )1 = tv r` .
• k = 2:
(tv r` )2 = tv r` tv r`
= tv tv r` r`
= tv tv 1
= t2v
where r` (v) = v, because v is parallel to and on top of `, and thus, r` tv = tr` (v) r` =
tv r` ..
• k = 3: (tv r` )3 = (tz r`2 (tv r` ) = t2v tv r` = t2v r` .
• k = 4: (tv r` )4 = (tv r` )2 (tv r` )2 = t2v t2v = t4v .
2. On the other hand, we will show that the rotational subgroup of the point group can be
realized as a subgroup of G, in a very precise sense. (So what happened in the previous
problem for reflections is all that goes wrong, in some sense, in trying to realize the point
group as a subgroup of G).
Let G < M2 be a subgroup and let ḡ ∈ im(π|G ) be a nontrivial rotation. Let g ∈ G be
any element such that π(g) = ḡ. Show that g is also a rotation, by the same angle as
ḡ (e.g., if g is a rotation matrix by 30 degrees counterclockwise, then g must also be a
rotation by 30 degrees counterclockwise about some point). Conclude that the rotational
subgroup Ḡ0 := ker(det |Ḡ ) is the isomorphic image of a subgroup G0 < G, i.e., there exists
a subgroup G0 < G such that π|G0 is an isomorphism π|G0 : G0 → Ḡ0 . (Such a subgroup
G0 is called a (group) section of Ḡ0 under the homomorphism π.)
1
Hints: To see that g is also a rotation, use Theorem 6.3.4 (consider the image under π
of elements in Theorem 6.3.4). To find G0 , observe first that Ḡ0 is cyclic (use Theorem
6.4.1), say with generator ḡ, and if g is such that π(g) = ḡ, conclude that G0 := hgi has
the desired property.
3. Chapter 6, Exercise 5.1. Let `1 and `2 be lines through the origin in R2 that intersect in
an angle π/n, and let ri be the reflection about `i and prove that r1 and r2 generate the
dihedral group Dn .
First, consider what happens when we reflect about both lines, using the rules from page
160 of Artin and the fact that reflecting about a line at some angle from the original line
is the same as rotating by the negative of that angle (to bring the second line on top of
where the original was), reflecting about the original line, and rotating back by the same
angle:
r`2 r`1 = ρ nπ r`1 ρ− nπ r`1
= ρ nπ r`1 r`1 ρ nπ
= ρ nπ 1ρ nπ
= ρ nπ ρ nπ
= ρ 2π
n
And we can see that hρ 2π i = {id, ρ 2π , . . . , ρ 2(n−1)π }, and hρ 2π i = n. General elements
n
n
n
n
can be represented in the form ρ 2πk . There are four cases of composing these elements
n
with the reflections:
• ρ 2πk r`2 :
n
ρ 2πk r`2 = ρ 2π(k+1) ρ− 2π r`2
n
n
n
= ρ 2π(k+1) ρ
n
π
−n
ρ −π r`2
n
= ρ 2π(k+1) ρ− nπ r`2 ρ nπ
n
= ρ 2π(k+1) r`1
n
• r`2 ρ 2πk = ρ −2πk r`2 .
n
n
• r`1 ρ 2πk = ρ −2πk r`1 .
n
n
• ρ −2πk r`1 . In the above cases, let k = k + 1, k = −k, and k = −k respectively, and
n
we see that, no matter what, any composition of one of the rotations and a reflection
will lead to an element of this form.
Thus, we see that every element can be formed from r`1 and ρ 2π , meaning that we have
n
hr`1 , ρ 2π i = Dn as our symmetry group, as seen many times before.
n
2
4. Chapter 6, Exercise 5.2. What is the crystallographic restriction for a discrete group of
isometries whose translation group L has the form Za with a 6= 0?
By the definition of Za, a 6= 0, all translational symmetries must be a multiple of a,
and this means we are working where there is only one translation. Essentially, there is
repetition only in one direction, meaning that our discrete group of isometries is a frieze
group. The crystallographic restriction tells us to consider the point groups which may
occur in this case. We know that all rotations must also be symmetries, and therefore,
send each translation vector to another. Since all translation vectors must be multiples
of only a single translation vector, they must lie along a single line, and therefore, the
rotation must send this line to itself. The only rotations which do this are of an angle π or
2π = id with rotational center on the line. Similarly, the only reflections which send this
line to itself are those perpendicular to the line, meaning that there is only one of them, up
to translation. Including neither gives the point group C1 . Including a rotation by π gives
us C2 , and adding the reflection to either of these gives us D1 and D2 respectively, which
must therefore be the only possible point groups of isometries whose translation group L
has the form Za, with a 6= 0.
5. Chapter 6, Exercise 5.6.
Let G be the group of symmetries of the frieze pattern
. Determine the point group G of G and the index in G of the
subgroup of translations.
The frieze pattern has reflections about what I can only see as the space between two
goal posts, either facing up or facing downward, and has rotational symmetry about the
points where the symmetry lines intersect the horizontal line through the middle of the
frieze pattern (which is a line of reflection for the only glide symmetry, with length equal
to the distance between any two consecutive vertical reflection lines) with angle π, as well
as about the points midway between them, all as shown here:
This makes G = D2 .
.
The subgroup of translations is simply all horizontal translations by a multiple of the
distance between any pair of “goalposts”, and there are 4 cosets of it: it, its rotation by π,
its reflection about one of the vertical symmetry lines, and its reflection about the same
line followed by a rotation through one of the rotations points by π.
6. Chapter 6, Exercise 6.1.
(a) Determine the point group G for each of the patterns depicted below...
1.
There are no rotations, reflections, or glide reflections, so the point group is G = C1 .
3
2.
There is a rotation of π about any of the points where four of the enclosed polygons
meet, and no reflectional or glide symmetries. Thus, G = C2 .
3.
There is a rotation of 2π
3 about any point where three of the polygons meet, and
no reflectional or glide symmetries. Thus, G = C3 .
4.
There is a rotation of π2 about any point where four of the polygons meet, or the
center of any polygon, and no reflectional or glide symmetries. Thus, G = C4 .
5.
There is a rotation of π3 about the center of any of the pinwheel polygons, but no
reflectional or glide symmetries, and thus G = C6 .
6.
4
There is a reflectional symmetry about any vertical axis passing through any vertical line between polygons, or through the point of any arrow, but no rotational
or glide symmetries (except those about the line of reflection). Therefore, G = D1 .
7.
There is no reflectional or rotational symmetry, but there is a glide symmetry about
the vertical line of reflection shown below, followed by a shift up or down of the
height of one of the wedges at the end of the polygon which are “cut off” by the
line of reflection.
Thus, G = D1 .
8.
There is a reflectional symmetry about any vertical line through the vertical joining
line between any two “scales” in the figure. There are no rotational symmetries,
but there is a glide symmetry about any vertical line halfway between any two
reflectional symmetry lines, some of which are not reflectional symmetry lines,
and this is what makes it unique from figure 6. Nevertheless, since it only has
reflections, the point group is G = D1 .
9.
5
There are reflectional symmetries about any vertical line through the tips of the
diamonds, as well as any horizontal line with the same property. It has rotational
symmetry with angle π about any joining of four diamonds or the center of any
diamond. These cover all glide symmetries as well. Thus, G = D2 .
10.
There are no reflectional symmetries in this figure, but there is a rotational symmetry of angle π2 about the point shown below, as well as all corresponding points:
as well as the center of the bricks (should have realized that before making the
picture) and there are glide reflections with vertical lines of reflection as shown
below, followed by a translation up or down by the height necessary to bring one
“point” that is reflected across to the location of the current point above it, where
it will sit and be happy:
This means that G = D2 .
11.
There is a rotational symmetry of
π
2
about the points such as the one shown below:
6
and reflectional symmetry about any vertical line which passes through the “point”
of each of the lines which wiggles across the figure. There is also a glide reflection
about the horizontal line passing midway between two “wiggles” with translation
vector left or right of length equal to the width of one “wiggle”. Thus, G = D2 .
12.
This “brick wall” has rotational symmetry of π2 about the center of any brick, as
well as through the point on the edge of each layer of bricks midway between the
edge of one brick and the edge of the next brick above/below it. There is reflectional
symmetry about the vertical line through the edges of bricks on alternating layers,
and with horizontal lines passing through the middle of any layer of bricks, not
touching the edges. There is a glide symmetry in the vertical direction through the
second set of rotation points, with translation up or down by the height of a single
brick, as well as glide symmetry by any horizontal line passing through between
layers of bricks, followed by a translation left or right by half the length of a brick.
Thus, G = D2 .
13.
This figure has rotational symmetry with angle π2 about the center of each circle as
well as the points where any four circles touch in tangential pairs, and reflectional
symmetry through the vertical lines passing through both the center of the circles
and the corresponding tangential meeting points of the other circles, as well as the
horizontal lines with the same property. There are two more reflections, with lines
of reflection at an angle of π2 to the horizontal/vertical passing through the centers
of circles only, or through tangential intersection points only. All glide reflections
are contained within this, so G = D4 .
14.
7
This figure has rotational symmetry with angle
π
2
about the points as shown:
and reflectional symmetry about the vertical lines through the points not selected
above (the other intersections of four polygons), and the corresponding horizontal
lines, as well as glide reflections about the lines which make angle π2 with the
vertical/horizontal, that mostly passes through the edges, such as the one shown
below:
followed by a translation in the same direction by the length of one of the mostly
edges of the polygons, exactly 4 of which are covered by the line segment shown
above. These 4 reflections/rotations mean that G = D4 .
15. THIS FIGURE HAS AN ERROR IN IT. THE CIRCLED LINE SEGMENT (AND
ITS EXTENSION) SHOULD NOT BE IN THE FIGURE.
8
The corrected figure, which I will use for this problem, is shown below:
This figure has rotational symmetry about the center of any of the hexagons or
tri-hexagonal figures with angle 2π
3 , and reflectional symmetry about any vertical
line that passes through the line between two neighboring tri-hexagonal figures and
bisects the hexagon between them, and the tri-hexagonal figure above that. There
is also reflectional symmetry about the lines which make an angle of π3 with the
vertical/ π6 with the vertical and pass through the same places as the above listed
lines. Thus, G = D3 .
16.
This figure has rotational symmetry with angle 2π
3 about any of the points where
the tips of 6 of the fleur-de-lis like figures and also about the point where any three
of the “leaves” meet. There is reflectional symmetry along the length of any fleurde-lis, which gives three reflectional lines: vertical, and two which make angles of
π
3 with it. This means that G = D3 .
17.
9
This figure has all the symmetries. About the center of any circle, which is also the
tangential intersection point of 6 other circles, there is 6-fold rotational symmetry–
that is, symmetry with angle π3 . In addition, there is reflectional symmetry about
the centers of the circles in the vertical direction, and the two lines which make
an angle of π3 with the vertical and also pass through the centers of the circles.
All of these lines pall through the squished “football” shapes exclusively. There
are three other reflectional symmetries, through the horizontal and the lines which
make angles of π3 with it, all of which pass through the centers of circles, but pass
through the squished triangles, and bisect the squished footballs shot-ways. Thus,
at long last, we have G = D6 .
(b) For which of the patterns can coordinates be chosen so that the group G operates on
the lattice L. Hint: Fix an origin. Consider what happens to the lattice under a glide
reflection, and under a rotation about the origin. You may use the results of the first
two problems if you like.
7. Chapter 6, Exercise 6.3. With each of the patterns shown, determine the point group and
find a pattern with the same type of symmetry in table 6.6.2.
1.
This figure also has a mistake. It should include the lines in red shown below, whose
absence probably caused me to waste a good 40 minutes on this problem:
This figure has four-fold rotational symmetry (angle π2 ) about the center of any square,
or the point where any four rectangular “bricks” meet. It has glide symmetry along
lines which make an angle of π2 with the horizontal and pass through the corners of a
square, as shown below:
10
with translation vector of the length of the segment shown above. There is a corresponding glide reflection along the line perpendicular to the one shown (as mentioned
above), and another pair such as the one illustrated below:
Where the line drawn shows both the direction of the line of reflection and its length.
There is a corresponding vertical glide symmetry. This makes the point group G = D4 .
I cannot find any of the other figures whose point group is D4 and which only have
rotational and glide symmetry.
2.
This figure has rotation with angle π2 about the center of any square, as well as rotation
with angle π about the center of any rectangle. It has horizontal and vertical lines of
reflection about the center of any strip of rectangles in the correct direction. Thus,
G = D4 . Since it has all the rotations and reflections, it has the same type of symmetry
as number 13:
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3.
This figure has rotational symmetry with angle π about the center of any star (inside
six hexagons), and rotational symmetry with angle π2 about the center of any of the
squares, such as the one shown below:
It has vertical reflectional symmetry about the middle of any star, as well as horizontal.
This *should* have another glide reflection at an angle of π2 , but I can’t find the darn
thing anywhere.
Regardless, the rotational symmetry should make the point group G = D4 . It does, I
believe, have the same symmetry pattern as number 14:
4.
12
This figure has no reflectional, rotational, or glide symmetries, and thus it must be
G = C1 , and the same symmetry as figure 1 above:
8. Chapter 6, Exercise 9.2. Let G be the group of rotational symmetries of a cube, let
Gv , Ge , Gf be the stabilizers of a vertex v, an edge e, and a face f of the cube, and let
V, E, F be the sets of vertices, edges, and faces respectively. Determine the formulas that
represent the decomposition of each of the three sets into orbits for each of the subgroups.
I will label the vertices of the cube as follows: (This was fun to draw in LaTeX)
2
1
3
4
6
5
7
8
Now G = {1, Pv(1 7)−120◦ , Pv(1 7)−240◦ , Pv(2 8)−120◦ , Pv(2 8)−240◦ , Pv(3 5)−120◦ , Pv(3 5)−240◦ ,
Pe(1 5−3 7)−180◦ , Pe(1 2−7 8)−180◦ , Pe(2 6−4 8)−180◦ , Pe(3 4−5 6)−180◦ , Pe(1 4−6 7)−180◦ ,
Pe(2 3−5 8)−180◦ , Pf (1 4 8 5−2 3 7 6)−90◦ , Pf (1 4 8 5−2 3 7 6)−180◦ , Pf (1 4 8 5−2 3 7 6)−270◦ ,
Pf (1 2 6 5−4 3 7 8)−90◦ , Pf (1 2 6 5−4 3 7 8)−180◦ , Pf (1 2 6 5−4 3 7 8)−270◦ , Pf (1 2 3 4−5 6 7 8)−90◦ ,
Pf (1 2 3 4−5 6 7 8)−180◦ , Pf (1 2 3 4−5 6 7 8)−270◦ }, where Pv (a b) are rotations about the space
diagonal between a and b, Pe(a b−c d) is a rotation about the line from the midpoint of
line segment ab to the midpoint of the line segment cd, and Pf (a b c d−e f g h) is a rotation about the line from the midpoint of face abcd to the midpoint of face ef gh. Let
vertex 1 be v1 , edge 1 4 be e1 , and face 1 2 3 4 be f1 . These are the vertex, edge,
and face where we will find the stabilizers. Note that V = {1, 2, 3, 4, 5, 6, 7, 8}, E =
{1 2, 2 3, 3 4, 4 1, 5 6, 6 7, 7 8, 8 1, 1 5, 2 6, 3 7, 4 8}, and F = {1 2 3 4, 1 2 6 5, 2 3 7 6, 3 4 8 7,
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1 4 8 5, 5 6 7 8}. Then we see that the rotations that will fix vertex v1 are those with a
pole through it–namely, the space diagonal through 1 7. Then the subgroup that stabilizes
the vertex are Gv1 = {g ∈ G|gv1 = v1 } = {1, Pv(1 7)−120◦ , Pv(1 7)−240◦ }. Similarly, the only
rotations which will fix edge e1 are those through the point of the middle of the edge and
its opposite, so Ge1 = {g ∈ G|ge1 = e1 } = {1, Pe(1 4−6 7)−180◦ }. Finally, the stabilizers
of the face f1 are going to be the rotations through the point in the middle of the face,
thus Gf1 = {1, Pf (1 2 3 4−5 6 7 8)−90◦ , Pf (1 2 3 4−5 6 7 8)−180◦ , Pf (1 2 3 4−5 6 7 8)−270◦ }. Now
I will show the decomposition of each of the three sets into orbits for each of these three
subgroups:
• Gv1 :
– V : The rotations send 1 to itself: O1 = {1}. They will send 2 to 4 and 5, so
O2 = {2, 4, 5}. They send 3 to 8 to 6, so O3 = {3, 8, 6}, and they send 7 to itself,
so O7 = {7}. In the interest of brevity and saving trees, I will simply list the
orbits in the rest of this problem.
– E: O1 2 = {1 2, 1 4, 1 5}, O2 3 = {2 3, 4 8, 5 6}, O4 3 = {4 3, 5 8, 2 6}, O3 7 =
{3 7, 8 7, 6 7}.
– F : O1 2 3 4 = {1 2 3 4, 1 4 8 5, 1 5 6 2}, O5 6 7 8 = {5 6 7 8, 2 3 7 6, 4 8 7 3}.
• Ge1 :
– V : O1 = {1, 4}, O2 = {2, 8}, O3 = {3, 5}, O6 = {6 7}.
– E: O1 4 = {1 4}, O1 2 = {1 2, 4 8}, O1 5 = {1 5, 4 3}, O3 7 = {3 7, 5 6}, O2 6 =
{2 6, 87}, O6 7 = {6 7}.
– F : O1 2 3 4 = {1 2 3 4, 4 1 8 5}, O3 4 8 7 = {3 4 8 7, 5 1 2 6}, O5 6 7 8 =
{5 6 7 8, 3 7 6 2}.
• G f1 :
– V : O1 = {1 2 3 4}, O5 = {5 6 7 8}.
– E: O1 2 = {1 2, 2 3, 3 4, 4 1}, O1 5 = {1 5, 2 6, 3 7, 4 8}, O5 6 = {5 6, 6 7, 7 8, 8 1}.
– F : O1 2 3 4 = {1 2 3 4}, O1 2 6 5 = {1 2 6 5, 2 3 7 63 4 8 7, 4 1 5 8}, O5 6 7 8 =
{5 6 7 8}.
9. Chapter 6, Exercise 9.6. Identify the groups of symmetries of a baseball, taking the seam
(but not the stitching) into account and allowing orientation-reversing symmetries.
There is a tennis ball with this homework. Its name is BB-8. It has the same seam pattern
as a baseball, without the stitching, and so has the symmetries which this question is
requesting. (Ignore the logo.) Enjoy!
For posterity’s sake, I will list them here anyway: There are two reflectional axes of
symmetry, both of which pass directly between one section where the seams approach eachother closest, parallel to that approach, and around to bisect the seam on its other close
approach. I have drawn both of these as great circles completely around the tennis ball.
There are three axes of rotational symmetry with angle of rotation 180◦ . The first, which
I denoted by x1 on the ball, passes directly through the points where the rotational axes
intersect–that is, the point in the middle of where the seams make their closest approach.
The other two axes, x2 and x3 are at the points where the seam changes concavity between
closest approaches. Again, all of these are labelled on the provided tennis ball, with both
14
ends of the axis labelled. Simply take the ball and place fingers on both points labelled
with the same value, and spin 180◦ to see the symmetry.
As was discussed in office hours, this is D4 D2 (The professor wasn’t 100% clear about his
notation.).
10. Chapter 6, Exercise 12.2. Describe the orbits of poles for the group of rotations of an
octahedron.
I shall label the octahedron as shown below: (Couldn’t find one at the store, sorry.)
1
6
3
4
5
2
The poles go through any vertex, the midpoint of any edge, and the center of any face
(which is well defined since they are equilateral triangles), and thus there are 24 of them,
6 vertices, 12 edges, and 8 faces. A rotation by 90◦ , 180◦ , or 270◦ about any vertex-pole,
rotation by 180◦ about any edge-pole, and rotation by 120◦ or 240◦ about any face-pole is a
symmetry. The vertices, edges, faces, and rotations shall be labelled in the same way as they
were for the cube in problem 8. For example, Pv(4)−−90◦ is a rotation about the axis through
vertex 4–and thus 3, of 90◦ in a clockwise direction as viewed from the vertex.
First, I shall consider the orbit for an arbitrary vertex-pole, say 1. id(1) = 1, Pv(4)−90◦ (1) =
6, Pv(4)−180◦ (1) = 2, Pv(4)−270◦ (1) = 5, Pv(5)−90◦ (1) = 4,Pv(5)−270◦ (1) = 3, and so we can see
that the pole, chosen arbitrarily, can be sent under symmetry to all other vertices, and thus
for any vertex-pole v, Ov = {1, 2, 3, 4, 5, 6, }.
Now, consider an arbitrary edge-pole. I shall choose 1 4. id(1 4) = 1 4, Pf (1 4 5)−120◦ (1 4) =
4 5, Pf (1 4 5)−240◦ (1 4) = 5 1, Pf (1 6 4)−120◦ (1 4) = 6 1, Pf (1 6 4)−240◦ (1 4) = 4 6, Pv(5)−90◦ (1 4) =
4 2, Pv(5)−180◦ (1 4) = 2 3, Pv(5)−270◦ (1 4) = 3 1, Pv(4)−270◦ (1 4) = 5 4, Pe(5 4)−180◦ (1 4) =
2 5, Pe(1 6)−180◦ (1 4) = 6 3, Pe(5 4)−180◦ (1 4) = 3 5. Thus, the edge-pole can orbit to
any other edge-pole, and thus they are in the same orbit, and for any edge-pole e, Oe =
{1 4, 1 5, 1 3, 1 6, 3 6, 6 4, 4 5, 5 3, 2 3, 2 6, 2 4, 2 5}.
Finally, consider an arbitrary face-pole. I shall chose 1 4 5. id(1 4 5) = 1 4 5, Pv(1)−90◦ (1 4 5) =
1 5 3, Pv(1)−180◦ (1 4 5) = 1 3 6, Pv(1)−270◦ (1 4 5) = 1 6 4, Pv(4)−180◦ (1 4 5) = 2 4 6, Pv(4)−270◦ (1 4 5) =
2 4 5, Pv(5)−180◦ (1 4 5) = 2 3 5, Pe(3 5)−180◦ (1 4 5) = 2 6 3. Again, we can see that any face-pole
can go to any other face-pole, and therefore, for face-pole f , Of = {1 4 5, 1 5 3, 1 3 6, 1 6 4, 2 4 5,
5 3, 2 3 6, 2 6 4}.
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