Prove Perfect Lie Algebra of R^3 Euclidean Motions Isn't Semisimple

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Does anybody know the answer of the following problem?

Show that the Lie group of Euclidean motions of R^3 has a Lie algebra g which is perfect i.e., Dg=g but g is not semisimple.
By Dg I mean the commutator [g,g] and a semisimple lie algebra is one has no nonzero solvable ideals.

Regards
 
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Is the following helpful?
1. Write down the lie algebra
2. show that Dg=g
3. find a solvable non-zero ideal (there will be an obvious candidate - this must be the easiest part)
 
does the lie group of Euclidean motions of R^3 means O(3) i.e. rotations?
 
No. I don't think so. I think it means distance preserving maps (not just linear maps) - so rotations and translations, and reflections. (SO(3) is the group of rotations, not O(3)).
 
you're right. So what is the Lie algebra of transformation preserving distance?
Since I'm learning Lie theory by myself, I have lots of questions even in definitions!
 
Write down the lie group, then work out its lie algebra. I'll give you the lie group.

G:= O(3)\sdp R^3, that is the semi direct product. It is the set of pairs (X,V) with X in O(3) and v in R^3, and the composition:

(X,v)(Y,w)=(XY,Xw+ v).

You have been told how to find lie algebras (or someone else has in another thread currently active in this forum).
 
Ok!

many thanks
 

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