- #36
cianfa72
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Another point: since SE(n) is not commutative then if we pick a=tr do exist t' and r' such that a=r't' ?
Vanadium 50 said:This is what makes parallel parking possible.
Yes.cianfa72 said:Another point: since SE(n) is not commutative then if we pick a=tr do exist t' and r' such that a=r't' ?
In fact, if R is the subgroup of rotations about a specific point, you must take r’ = r.mathwonk said:you can even take r' = r.
You mean the set ##\{ r^{-1}tr, t \in T \}## is a subgroup and is equal to T itself.mathwonk said:so if r is any rotation, the subgroup r^-1Tr is again equal to T.
Yes. This is the definition of a normal subgroup. That this holds regardless of what ##r## is.cianfa72 said:You mean the set ##\{ r^{-1}tr, t \in T \}## is a subgroup and is equal to T itself.
cianfa72 said:Another point: since SE(n) is not commutative then if we pick a=tr do exist t' and r' such that a=r't' ?