3 correct spelling

Suppose one has a convex 4-gon in the plane. What symmetry groups can it have?

The graph of this convex 4-gon is a 4-cycle so as a graph its automorphism group is the dihedral group which I will denote D(4) which has 8 elements. Now there is a convex 2-dimensional polygon which has this as its group, namely a square. However, The group D(4) has a cyclic subgroup of order 4, yet there is NO convex 4-gon which has the cyclic group of order 4 as its set of isometries. There is a rectangle with unequal sides which has a group of order 4 as its symmetry group but this is the Klein group, not the cyclic group, of order 4.

For 3-dimensions, a similar thing can happen. It is known that the vertex-edge graph of any 3-dimensional convex polytope is a planar and 3-connected graph and the converse holds. This is Steinitz's Theorem. Suppose H is such a graph (e.g. planar and 3-connected) and the (full) automorphism group of H is G. There is a beautiful theorem of Peter Mani's which states H can be realized in 3-space by a metric polytope P which has group G as its group of isometries. However, it does not follow that for any subgroup I of G that there is a 3-dimensional convex polyhedron whose group of isometries is I. In fact, for the group with 48 elements which is the isometry group of the graph of the 3-cube there is a subgroup of order 24, the rotation group, but there is no 3-dimensional convex polyhedron which is combinatorially a 3-cube which has 24 isometries.

Here is the reference for Mani's paper:

P. Mani, Automorphismen von polyedrischen Graphen, Math. Ann. 192 (1971) 279–303.

This is a generalization of this theorem to complexes, as mentioned in this paper:

http://arxiv.org/abs/math/0310165

If you restrict your attention to graphs rather than polytopes there is a nice theorem of Roberto Frucht.

For any finite group H, there is a graph G(H) such that the automorphism group of G(H) is H.

There are extensions of Frucht's Theorem including to 3-valent graphs. If there was an extension of Fruckht's Frucht's theorem to planar 3-connected graphs than via Steinitz's Theorem the original question would be answered. I am not sure if this has been done or not.

A survey paper (about graphs with specified automorphism groups and related matters) of Babai's is available:

www.cs.uchicago.edu/files/tr_authentic/TR-94-10.ps

2 added 889 characters in body

Suppose one has a convex 4-gon in the plane. What symmetry groups can it have?

The graph of this convex 4-gon is a 4-cycle so as a graph its automorphism group is the dihedral group which I will denote D(4) which has 8 elements. Now there is a convex 2-dimensional polygon which has this as its group, namely a square. However, The group D(4) has a cyclic subgroup of order 4, yet there is NO convex 4-gon which has the cyclic group of order 4 as its set of isometries. There is a rectangle with unequal sides which has a group of order 4 as its symmetry group but this is the Klein group, not the cyclic group, of order 4.

For 3-dimensions, a similar thing can happen. It is known that the vertex-edge graph of any 3-dimensional convex polytope is a planar and 3-connected graph and the converse holds. This is Steinitz's Theorem. Suppose H is such a graph (e.g. planar and 3-connected) and the (full) automorphism group of H is G. There is a beautiful theorem of Peter Mani's which states H can be realized in 3-space by a metric polytope P which has group G as its group of isometries. However, it does not follow that for any subgroup I of G that there is a 3-dimensional convex polyhedron whose group of isometries is I. In fact, for the group with 48 elements which is the isometry group of the graph of the 3-cube there is a subgroup of order 24, the rotation group, but there is no 3-dimensional convex polyhedron which is combinatorially a 3-cube which has 24 isometries.

Here is the reference for Mani's paper:

P. Mani, Automorphismen von polyedrischen Graphen, Math. Ann. 192 (1971) 279–303.

This is a generalization of this theorem to complexes, as mentioned in this paper:

http://arxiv.org/abs/math/0310165

If you restrict your attention to graphs rather than polytopes there is a nice theorem of Roberto Frucht.

For any finite group H, there is a graph G(H) such that the automorphism group of G(H) is H.

There are extensions of Frucht's Theorem including to 3-valent graphs. If there was an extension of Fruckht's theorem to planar 3-connected graphs than via Steinitz's Theorem the original question would be answered. I am not sure if this has been done or not.

A survey paper (about graphs with specified automorphism groups and related matters) of Babai's is available:

www.cs.uchicago.edu/files/tr_authentic/TR-94-10.ps

1

Suppose one has a convex 4-gon in the plane. What symmetry groups can it have?

The graph of this convex 4-gon is a 4-cycle so as a graph its automorphism group is the dihedral group which I will denote D(4) which has 8 elements. Now there is a convex 2-dimensional polygon which has this as its group, namely a square. However, The group D(4) has a cyclic subgroup of order 4, yet there is NO convex 4-gon which has the cyclic group of order 4 as its set of isometries. There is a rectangle with unequal sides which has a group of order 4 as its symmetry group but this is the Klein group, not the cyclic group, of order 4.

For 3-dimensions, a similar thing can happen. It is known that the vertex-edge graph of any 3-dimensional convex polytope is a planar and 3-connected graph and the converse holds. This is Steinitz's Theorem. Suppose H is such a graph (e.g. planar and 3-connected) and the (full) automorphism group of H is G. There is a beautiful theorem of Peter Mani's which states H can be realized in 3-space by a metric polytope P which has group G as its group of isometries. However, it does not follow that for any subgroup I of G that there is a 3-dimensional convex polyhedron whose group of isometries is I. In fact, for the group with 48 elements which is the isometry group of the graph of the 3-cube there is a subgroup of order 24, the rotation group, but there is no 3-dimensional convex polyhedron which is combinatorially a 3-cube which has 24 isometries.