Hey there everyone,
Does D3 (The dihedral group of order 6) is indecomposable?
Proof is needed...
TNX!
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...
Well... D3 isn't commutative indeed... Each subgroup of order 2 of D3 is of the form
{1,a} where a is a reflection and it's commutative...
I can't think of an example of a subgroup of order 3 so I can't answer your question completely....If the subgroup of order 3 of D3 is also not commutative, so because of the isomorphism between the two, we will get a contradiction..But what are the subgroups of order 3 of D3?
TNX
As you said, $\displaystyle D_3$ is non-commutative. However, if it was decomposable when what would it look like? It would have to be the direct product of $\displaystyle C_2$ and $\displaystyle C_3$, the cyclic groups of order 2 and 3 respectively.
Is this group abelian?