Recent content by Daettil

Yes the group will have order 10. If you label a vertex A and label a vertex adjacent to A B. Then there are 5 positions that A can be moved to and each of those allows exactly 2 positions for B.

If you center your pentagram about the origin then the reflection across the x-axis would be a reflectional symmetry. You should be able to generate your group from that reflection and the rotation of 72 degrees about the origin.

The rotational symmetries are isomorphic to the cyclic group of order 5, but there are also reflectional symmetries that need to be considered.

Perhaps you should try to construct a subgroup of order 30. Such a subgroup is normal so you can constuct one by taking the union of conjugacy classes...

Well Matrix multiplication is associative so you should be able to manipulate (AB)C to get C(AB) pretty easily.

Try taking 3 elements from G, say g, h, and gh. You should be able to show gh=hg pretty easily.