Homework #6 Solutions
5.6.1 What is the stabilizer of the coset aH for the operation of G
on G/H?
Answer: The stabilizer of aH is the conjugate subgroup aHa-1.
Proof:
By definition, the stabilizer of aH is GaH={g in G : g(aH)=aH}.
But gaH=aH ó a-1gaH=H (via left multiplication of both sides by a-1 to get “=>”, or by a to get , “<=”),
and a-1gaH=H ó a-1ga in H ó g in aHa-1.
5.7.2 Let G be the
group of rotational symmetries of a cube C. Two regular tetrahedral Δ,
Δ’ can be inscribed in C, each using half the vertices. What is the order
of the stabilizer of Δ?
The red vertices, when connected by edges, are the vertices of a regular tetrahedron Δ inscribed in the cube. This is because all the sides are congruent, of length equal to that of the diagonal of the side of the cube. The blue complementary vertices are also the vertices of a regular tetrahedron Δ’ inscribed in the cube.
The group G of rotational symmetries of the cube acts on the set {Δ, Δ’} of the two tetrahedrons. This is because any rotational symmetry f in G either preserves the two tetrahedra (if f=identity map or of f=rotation by 120 or 240 about any of the vertices of C) or swaps Δ for Δ’ and viceversa (for example if f=rotation by 90 degrees about lines connecting midpoints of opposite sides). The orbit of either one of the tetrahedra is of size 2.
You can either list carefully all rotation symmetries and count those that stabilize Δ (and hence also stabilize Δ’, of course) or, much easier, use the counting formula:
|G|=|OΔ ||GΔ| => 24=2|GΔ| => |GΔ|=12.
Extra Problem #1:
Classify all groups G of order 6 up to
isomorphism.
a) What orders can the elements have?
Since the order of any element must divide the order of the group, G can only have elements of order 1, 2, 3, or 6. There’s always only one element of order 1, namely the identity. G has elements of order 6 if and only if it’s cyclic.
b) Prove that such a
group G must have at least an element x of order 3, by showing that it cannot
have all but the identity be of order 2.
If G is cyclic of order 6, then G has an element x of order 3 (the square of its generator).
If G is not cyclic, the only way it can have no element of order 3 is if all its non-trivial elements have order 2.
Suppose all elements g of G satisfy g2 =1. That is, G={1,a,b,c,d,e} with a, b,…, e elements of order 2. In particular, by the “Hello World” homework problem, G is abelian.
By closure, ab is in G. It cannot be either 1, a or b: ab≠1 because a is its own inverse and each element has a unique inverse. ab≠a because b≠1 and if ab=a we could cancel a and get b=1. Hence ab must be one of the other elements, say c=ab. By commutativity, it follows that ac=ca=b, and bc=cb=a. That is, {1,a,b,c} forms a subgroup of order 4 (isomorphic to the Klein group). This is impossible since 4 does not divide 6.
Hence any group G of order 6 must have an element x of order 3.
c) Prove that such a group G must
have at least an element y of order 2, by showing that it cannot have all but
the identity be of order 3.
Again, if G is cyclic then the cube of its generator is an element of order 2. If it’s not cyclic, then it has non-trivial elements of order 2 or 3 only.
If a is an element of order 3 in G then a generates a cyclic subgroup < a >={1, a, a 2} and any two distinct such subgroups must intersect trivially (since 3 is a prime). So we can have G={1, a, a2, b, b2, c}. But c cannot have order 3. If it did, there would have to be an element c2 in G, distinct from all the rest. This shows that not all nontrivial elements of G can have order 3, so there must be an element y of order 2.
d) Show that G is generated by x and y.
So far we have the following elements of G:
- 1
- An element x of order 3 (hence also x2)
- An element y of order 2.
By closure, G must also contain all the products of these elements.
By the same argument as before, xy and x2y must be distinct, new elements. Compare inverses or consider the results of cancellations to show:
xy ≠ 1, x, x2 or y ,
x2y ≠ 1, x, x2, y, xy.
Since we can get 6 distinct elements in G as products of x’s and y’s, x and y generate G.
e) Show that there are
only two possibilities for xy: either xy=yx (what group do you get?) or
xy=yx2(what group
do you get then?)
Consider now the product yx. It cannot equal 1, x, x2, or y, by the same sort of argument as before. So it must equal xy or x2y, since these are the only other elements in G.
If yx=xy, then x commutes with y. Since G is generated by two elements that commute with each other, G is abelian. By the structure theorem for finitely generated abelian groups, there is only one abelian group of order 6 (up to isomorphism): cyclic of order 6 (isomorphic to C2 x C3 ). [Alternatively, see that the relations x3=1, y2=1 and xy=yx uniquely determine all the products of any two elements in G, and the multiplication table you get for G is that for the cyclic group of order 6]
If yx= x2y (which is equivalent to xy=yx2), then G satisfies the definition of the dihedral 3-group D3: it is generated by two elements x and y that satisfy x3=1, y2=1 and xy= x2y. So G is isomorphic to D3. [Alternatively, check that the given relations uniquely determine the multiplication table for G and that you get the same table as for D3]
In conclusion: there are only two non-isomorphic groups of order 6: the cyclic one C6 and the dihedral one, D3.
Extra Problem #2: Consider the triangular prism in the picture (the ends are congruent equilateral triangles).
a) What are all the rotation
symmetries of this prism? What group structure do they
have?
There are three rotational symmetries that stabilize the triangular ends: 1, x, x2, where x corresponds to rotation by 120 degrees about the line connecting the centers of the faces of the two triangles, so x (and x2) has order 3.
The orbit of a triangular face has size two, because the face is either mapped to itself or to the other triangular face under the symmetries of the prism. There is a rotation by 180 degrees about the vertical line through the center of the prism that swaps the two faces. Call this y. Note that y has order 2.
Hence, by the counting formula, the group R of rotational symmetries has order 3x2=6. By the previous problem, R is either C6 or D3, depending on whether x and y commute or not. We can see if xy=yx by seeing if they map the vertices the same way.
:
We see that xy≠ yx, so R is isomorphic to D3 (which is also isomorphic to S3)
b) What are all the symmetries that
stabilize the end triangles? List them as elements of S3 by
considering their action on the triangle 123. What group do they form?
Besides 1,x, x2 described in part (a), there are also 3 reflections in planes that contain a vertex of each triangular face and the midpoint of the opposite sides. These all have order 2. Call one of them y’. The others are xy’ and x2y’. If we consider the action of these symmetries on one of the end triangles we see that we have a group isomorphic to D3.
c) What are all possible symmetries?
List them and identify the group they form in terms of familiar groups.
There are 6=| D3| symmetries that preserve an end triangle and the orbit of an end triangle is of size two. Hence there are 6x2=12 total symmetries.
Call the group of all symmetries G. Then G has order 12 and two distinct subgroup isomorphic to D3 which overlap, having the subgroup of order 3 in common.
Recall
the definition of elements x and y in part (a). Define c to be the following
symmetry: reflection in the blue middle plane:
Clearly c has order 2.
Check that G can be generated by x,y and c. (either by hard work, writing all 12 symmetries as products of x’s, y’s and c, or by a shortcut, showing that <x,y> intersects <c> trivially, so the set product <x,y><c> must have cardinality 6x2=12, so it must be the whole G)
Moreover, c commutes with either x or y from part (a) (verify via pictures, for example). Hence the group C={1,c} is a normal subgroup of G. But the subgroup H=<x,y> from part (a) (isomorphic to S3) is also normal (since it has index 2). And C intersects H trivially. Since the cardinalities of C and H multiply to 2x6=12=|G|, we can apply Proposition 2.8.6(c) and claim that G is isomorphic to H x C, i.e. to S3 x C2.
Remark: Notice that we could not use y’ from part (b) instead of the element c, because, while it’s still true that 1) G=<x,y><y’>,
2) and <x,y> intersects <y’> in {1} only,
3) and <x,y> is normal,
--- the subgroup <y’>={1,y’} is not normal so we cannot apply 2.8.6.