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For the case of the group <{I, R, R', R''}, *> of rotations of a square, I said:

|I| = 1

|R| = 4

|R'| = 2

|R''| = 4

For a rhombus, do I need to take into consideration the symmetry of it to obtain the order of its rotations? or can I say its rotations have the same order as in a square given that they have the same number of elements?

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  • $\begingroup$ What do you mean by diamond? The octahedron? $\endgroup$ Commented Nov 19, 2021 at 3:28
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    $\begingroup$ A diamond has fewer rotations than a square. $\endgroup$ Commented Nov 19, 2021 at 3:31
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    $\begingroup$ @3311411 Sorry, a rhombus. I didn't know the term in english and google translate said it was called a diamond $\endgroup$ Commented Nov 19, 2021 at 3:35
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    $\begingroup$ I don't think that thinking about a square is a helpful approach. You are better off just considering the symmetries of a rhombus separately. There are 4 vertices. Which ones can map to which via a rotation? This should help you write down the rotations of a rhombus. $\endgroup$ Commented Nov 19, 2021 at 4:11
  • $\begingroup$ @tkf thanks! I will take that approach $\endgroup$ Commented Nov 19, 2021 at 4:15

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