Question on conditions for commutativity of subgroups

[mnb96]

Hi,
it is known that given two subgroups $H\subset G$, and $K\subset G$ of some group G, then we have that:

1) H, K are normal subgroups of G
2) $H\cap K$ is trivial

are sufficient conditions for H and K to commute.
Moreover we have that:

H, K commute $\Rightarrow$ H, K are normal.

In fact, conditions 1) and 2) together are not necessary conditions for commutativity because there exist subgroups that are commutative but do not have trivial intersection (it is posible to find examples).

My question is: is it possible to keep condition 1) and instead replace only 2) with some weaker condition that would make 1),2) necessary and sufficient conditions for commutativity?

 

[DonAntonio]

Hi,
it is known that given two subgroups $H\subset G$, and $K\subset G$ of some group G, then we have that:

1) H, K are normal subgroups of G
2) $H\cap K$ is trivial

are sufficient conditions for H and K to commute.
Moreover we have that:

H, K commute $\Rightarrow$ H, K are normal.

This is false: $\,K=\{(1)\,,\,(12)\}\,\,,\,\,H=\{(1)\,,\,(34)\}\,$ are commuting subgroups of $\,S_4\,$ and

they're far from being normal. In fact, they even commute pointwise.

DonAntonio

In fact, conditions 1) and 2) together are not necessary conditions for commutativity because there exist subgroups that are commutative but do not have trivial intersection (it is posible to find examples).

My question is: is it possible to keep condition 1) and instead replace only 2) with some weaker condition that would make 1),2) necessary and sufficient conditions for commutativity?

 

[mnb96]

oh!
I think I forgot to say that G is not just some arbitrary group, but I am considering: $$G=HK$$ In such case it should be true that "H, K commute ==> H,K normal". My original post was intended to be formulated under this assumption: G=HK.