Group of Symmetries of a Rectangle a. Carefully describe the group of symmetries of a rectangle...

Group of Symmetries of a Rectangle

a. Carefully describe the group of symmetries of a rectangle Describe the types, the orders, and the structures of the groups and their elements. After clearly naming the elements in some way, provide tables for each group. Describe them as a group of permutations on the vertices.

b. Next, carefully describe each of these groups as subgroups of some permutation group. Be sure to provide reasons for your choices.

c. What are the POSSIBLE orders for any subgroups of each group? Explain.

d. Next, carefully describe all the subgroups of each of these groups. Be sure to provide information about the structure of each subgroup, their order, the order of their elements. Provide generator(s) where possible.

e. Answer these questions about each group described in part a making sure to give reasons: Are any of these groups cyclic? Are any of these groups abelian? Which groups are cyclic? Which groups are abelian? Are there subgroups of every possible order? Which subgroups (in each group among the different groups) are isomorphic? How do you know they are isomorphic or not?

Solutions

Expert Solution

We can make the group D4, dihedral group of order 8 by 90o rotation with respect to the center of mass and 1800 rotation with respect to a fixed diagonal of a rectangle.

We consider two elements a and b where a is described by 900 rotation of the rectangle with respect to the center of mass and b is described by 1800 rotation with respect to a fixed diagonal.

Thus we get that the order of a is 4 and the order of b is 2. These a and b together make the dihedral group D4 i.e., D4 = <a,b>

A=<a> , B=<b> , C=<a2> are three subgroup of D4. .Clearly A, B, C are the cyclic subgroup of D4 and hence they are commutative. B*C is a non-cyclic commutative subgroup of D4. D4 is the only non-commutative subgroup of D4 because D4 is the only subgroup of D4 of order 8. All other subgroups must be of the order of 1,2 or 4 and hence commutative.

2,4 8 are the multiples of 8. B, A, D4 are the subgroup of order 2,4,8 respectively

Here B and C are of order 2 and hence they are isomorphic. A and B*C are of order 4. But A is isomorphic to Z4 and B*C is isomorphic to K4. So A and B*C are not isomorphic.

Colby Messinger answered 1 year ago

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