It is well-known that any two cyclic groups of the same order (number of elements) are isomorphic. Consider two cyclic matrix group, one group generated by
3 by 3 matrix, namely A = [0, 2, 1; 3, 0, 2; 4, 3, 1] over the field of integer modulo 5, ℤ5, and another group
generated by a permutation matrix
P := P_{(1,2,3,4,5,6,7,8,9,10)} ∈ M10. We see that and
are of ten elements cyclic groups, they are isomorphic.
The elements in group
is orthogonal, because if P is a permutation matrix, then P⁻¹ = P^{T}, in contrast (A^{i})^{T} ∉ for some i. Why orthogonality is not stable under group isomorphism?
dear Wanicharpichat which type properties arew different for two isomorphic cyclic matrix groups.. Is it related to representations...
But the orthogonality related to inner product are u sure that group isomorphism enough for inner product. To preserve inner product nit must be an isometry betyween them.
This is not a new phenomenon! For instance, look into subgroups of the additve group of reals. Z is one and Q is another, bot are countable, but Q is dense while Z is discrete, very far far away from being dense!
When one speaks of an isomorphism, it is related to the structure; so, if we change the structure they may or may not remain to be isomorphic further. In your question, orthogonality is not a property of groups.
Dear Wiwat,
I think the most enlightening way to interpret your question as well as Peter's answer, is to use the language of categories. Without being too precise, a category is a collection of sets with a given structure, called objects, AND a collection of morphisms, which are applications which must respect the structure. An isomorphism is a bijective map f such that both f and inverse f are morphisms. For example: - in the category of topological spaces, morphisms are continuous maps, isomorphisms are bicontinuous bijections (=homeomorphisms) - in the category of groups, morphisms are the usual group homomorphisms - in the category of vector spaces over a given field, morphisms are the usual linear applications - to speak of "orthogonal operators", you must first introduce a given vector space with a (non degenerate) inner product, and define an orthogonal operator as a linear operator which respects the inner product [Peter, I think this is the Bourbaki terminology; an orthogonal operator is necessarily injective, hence in finite dimension, bijective; the orthogonal operators then constitute the "orthogonal group" ]. If you want additional "metric" properties, you must impose "positive definite" properties to your inner product, etc. If you take coordinates w.r.t. an orthonormal basis, then the orthogonal operator is represented by an "orthogonal matrix". But a change of coordinates must then transform an orthonormal basis into another one, hence it is not only a mere linear transformation. As Peter says, "orthogonality" is not a mere group property !
Dear Prof. Thong nguyen quang do,
Thanks a lot.
With best regards
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