Originally Posted by
WannaBe I tried to figure out what is the contradiction we get from assuming that it's decomposable...I first thought about normality...If D3 is simple, then it's not decomposable, as required... The proof is kind of a Lemma for another proof, so I would be delighted if you'll be able to be more specific...The orders of the groups it would be decomposed to would be 3&2 (or 6 and 1 but it's trivial)... D3 has a sub-group of order 2 and a sub-group of order 3...So I can't realy see the contradiction...
Tnx in advance...