이현녀 선생님의 답변입니다.
어떤 사람이 이렇게 질문을 했습니다.
I would like to know following solution of S_3 is
not decomposible
is correct or not ?
질문자의 생각입니다.
sol>
Since |S_3| = 6 ,
If S_3 is isomorphic to H x K (H,K are subgroup of G)
and H=/={1}=/=K,
then W.L.O.G it must be H=<(12)>, K=<(123)> or
H=<(23)>,K=<(123)> or
H=<(13)>, K=<(123)>.
Now,
in all above three cases,
since H x K ~= C_6 (cyclic group of order 6) ,
H x K is cyclic group but S_3 is not a cyclic group.
therefore, S_3 has
no subgroups H,K satisfying S_3 ~= H x K,
H=/={1}=/=K.
=> so, S_3 is
not decomposible.
I read this solution at my book but, I think this
solution is false
because above H is not a normal subgroup of G.
I
know the definition of decomposible group as follows :
Any group G is
decomposible iff there are normal subgroups H,K of G
satisfying
G~= H x
K , |H|=/=1=/=|K|
리플이 이렇게 달렸습니다.
That would be another way of seeing
that G is not indecomposable (it
would have to have a normal subgroup of
order 2). But there is nothing
wrong with the argument given: the product of
a cyclic group of order
2 and one of order 3 is necessarily cyclic of order
6. Since G is not cyclic, that s it.
그러나 사실 전 모르겠습니다. 현대대수를 접한지 오래되서..
다만 인터넷에 떠도는 질문과 답이 있길래 올립니다. 스스로 판단하셔요.. |
