Can we say more about this structure?

Question

Let $N$ be a subgroup of $G$, One can show that,

$N$ is normal in $G$ if and only if for all $xy\in N$, $yx\in N$.

Above proposion can be proved by elemantary methods.

I wonder the following;

Let $H$ be a subgroup of $G$ with following property,

If $xyz \in H$ then all possible product of $x,y,z$ in $H$, i.e $xzy,yxz,zyx..\in H$.

By considering above argument, we can directly say that $H$ is normal in $G$ but actually it is stronger than normality so I wonder whether such $H$ must be characteristic or not or what can we say about $H$ in that case ?

Answer 1

If $xyz \in H$ and $xzy \in H$ then $(xyz)^{-1}(xzy)=z^{-1} y^{-1} z y = [z,y] \in H$.

Since for every $y,z \in G$ there is some $x=(yz)^{-1}$ so that that $xyz \in H$, we get that every such $H$ contains $[G,G]$. Conversely if $H$ contains $[G,G]$, then looking at the abelian group $G/H$ one gets the condition.

Conjugation gives you the cyclic permutations, $xyz \to yzx \to zxy$, but the other permutations imply a stronger property, containing the derived subgroup.

Answer 2

If you just cycle the products, then is again the same answer: it's equivalent. But if you take all the ordered products, than is the same (passing to the quotient) of a group where if $xyz=1$ than the properties remain true permuting the factors. Is easy to prove that this implies that G is abelian. So you're detecting the subgroups H such that G/H is abelian. That is H contains the commutator group.