Group Theory Problem: Finite Index of Intersection of Conjugates of H

[AKG]

Let H be a finite-indexed infinite subgroup of an infinite group G. Suppose:

$$G = \bigcup _{i = 1} ^{k} g_i H$$

then

$$J = \bigcap _{i = 1} ^{k} g_i H g_i ^{-1}$$

is a normal subgroup of G and an intersection of all of the (finitely many) conjugates of H. Show that J has a finite index.

 

[vsage]

A finite subgroup inside an infinite group can have a finite index? I would like to know how this problem is proved also. Having a finite group partition an infinite group into a finite number of pieces is counterintuitive to me.

 

[fourier jr]

Having a finite group partition an infinite group into a finite number of pieces is counterintuitive to me.

the integers can be partitioned into finite subgroups. think of 2Z, 3Z, 5Z, etc. they're all isomorphic to integers mod 2, mod 3, mod 5, etc which are finite. not sure what to do with the actual problem though, but at least we've got examples to play with. maybe there's a typo?

 

[AKG]

No, a finite subgroup inside an infinite group cannot have a finite index, my wording was extremely poor. When I said that J was a "finite intersection" I didn't mean that it had finite order, I meant that it was an intersection of finitely many sets. Hope that clears up some of the confusion.

 

[AKG]

While I'm at it, here's another problem I could use some hints on:

Find a solid whose full symmetry group is isomorphic to $A_4 \times \mathbb{Z}_2$.

 

[NateTG]

While I'm at it, here's another problem I could use some hints on:

Find a solid whose full symmetry group is isomorphic to $A_4 \times \mathbb{Z}_2$.

Well, one thing to look at is what the order of that group is. That might give you some idea. You could also look at the order of the symetry groups of various solids.

 

[AKG]

Well, one thing to look at is what the order of that group is. That might give you some idea. You could also look at the order of the symetry groups of various solids.

I don't think it's that simple. For one, I don't have a list of orders of the symmetry groups for all possible solids. Second, the symmetry group of the tetrahedron has the same order as $A_4 \times \mathbb{Z}_2$ but the two groups are not isomorphic; the symmetry group of the tetrahedron is $S_4$.

 

[CarlB]

Dear AKG;

The rotational group of the tetrahedron is $$A_4$$. (For example, see:
http://math.ucr.edu/home/baez/twf_ascii/week155 .)

That means that there are a set of 3x3 rotation matrices which forms the group $$A_4$$. As it turns out, -1 is not in that group (that is, the tetrahedron does not have a center of symmetry). Consequently, the rotation group defined by the tetrahedron, along with the negatives of the group, has $$A_4 \times Z_2$$ symmetry.

To define the general solid with this symmetry, [that is, to produce a solid that possesses the rotational symmetry but no other symmetry] begin with a randomly shaped solid [that is, don't start with a sphere], more or less centered around the origin, and enforce the rotational symmetries by keeping only those portions of the solid that are closer to the origin than any of the symmetrized boundaries.

A computerized milling machine may come in handy. If you need one, I've got some for sale.

Carl

 

[AKG]

This will give me a shape that has rotational symmetry of at least $A_4$, but possibly greater. Also, this doesn't ensure that the full symmetry group of this object contains -I. How do I know this process won't just give me back the tetrahedron? How do I use this process in real life? Even if I could do this process, and get something with rotational symmetry precisely isomorphic to $A_4$, if -I is not a symmetry of it, and I do a similar process to ensure that -I is, then I will probably end up adding more rotational symmetries to the group.

Another approach is to start with something like a cube, and shave off a bit of the edges or corners in the same orbit, thereby restricting some of the available rotations. This doesn't seem to be working out for me though...

The problem I'm much more concerned with right now is the one in the original post.

 

[AKG]

http://caramel.oc.chemie.tu-darmstadt.de/lemmi/script/redirect.cgi?filename=http://caramel.oc.chemie.tu-darmstadt.de/lemmi/structures/organic_symmetry7.html tells us that the tetrahedron has symmetry group $T_d$ and other useful information in the table on the right.

 

[Hurkyl]

This seems to be an easier construction:

Pick a point that is not at the origin.

Get new points by applying group elements to that point.

Take the polyhedron with the constructed points as vertices. (i.e. the convex hull of those points)

(This, of course, assumes you've already represented your group as rotation matrices)

 

[AKG]

Hurkyl, that construction does seem easier. Here's a shape that appears to work. This could be constructed with your construction if we start with the front-right vertex (it would be at a position like (4,1,0)). We would choose our rotations of order 2 to be the half-turns about each axis, and the rotations of order 3 are the rotations through 120 degrees about the axes determined by (1,1,1), (-1,1,1), (1,-1,1) and (1,1,-1).

 

[CarlB]

This will give me a shape that has rotational symmetry of at least $A_4$, but possibly greater.

Choose a randomly oriented plane by a vector perpendicular to the center of the plane from the origin. By "randomly oriented", it is more than sufficient to choose intersections with the x, y, and z axes that are not commensurate. For example, let the plane be defined by the equation
$$\pi x + e y + z = 0$$.

Now symmetrize that vector with your rotation group to obtain a collection of planes. Let the solid be the stuff that is closest to the origin. In effect, the solid is the solution of the linear programming problem. The reason for choosing a random plane is that you want to eliminate any possiblity for a symmetry that is not contained in the rotation group. Otherwise you could end up with a solid as simple as a tetrahedron, which as you say, has too much symmetry.

As it turns out, I wrote a program that solves the linear programming problem in 3 dimensions for the various rotational groups that correspond to the crystal symmetries, and published on the web. I liked the appearance of the gyroidal crystal class and used it as the default crystal. The gyroid has S4 symmetry. What you need is the diploid, which has A4 symmetry, obtained by adding a center of symmetry (or a mirror plane) to the basic tetrahedron rotation group.

I believe that a drawing of a solid that will fulfill your needs can be seen rotating in the java applet on the bottom of the website at this link by clicking on the "class++" button four times, to obtain the "diploidal" class:
http://www.brannenworks.com

This is the crystal symmetry of pyrite, which crystallizes in the diploidal class:
http://mathworld.wolfram.com/Pyritohedron.html

Carl

 

[CarlB]

Oops. That plane equation should be $$\pi x + e y + z = 1$$. Having the RHS equal to zero would put the plane through the origin.

Carl

 

[matt grime]

did you get the frst one sorted?

the only thing think of on looking at it is to consider

$$g_jg_iHg_i^{-1}$$

where does that lead

 

[AKG]

Yes, I believe I have the first one sorted. The point is to show that the index of J in H is no more than the index of H in J. Suppose x and y are in H. Let H' be the conjugation of H by some g in G. If xH' = yH', then xJ = yJ, where J is just the intersection of H and H' for now (and the proof should proceed by induction on the number of conjugates of H). So the number of distinct cosets of J in H can at most be the number of distinct cosets of H' in G, i.e. the number of cosets of J in H is at most the index of H, so J has an index of at most k², where k is the index of H, hence J has finite index.