Does the closedness of a subgroup $H$ of $G$ imply that for $h\in H, g\in G \setminus H : hg \not \in H$?

Question

Does the closedness of a subgroup $H$ of $G$ imply that for $h\in H, g\in G \setminus H: hg \not \in H$?

I'm new to abstract algebra and this would appear to be a useful result. However, by definition, a subgroup is nearly closed, and I don't think this directly implies that for any element $g$ from $G$ not in $H$, the composition $gh \not \in H$.

Is this in fact a true statement?

Answer 1

For any subset $A$ of $G$ that is closed under the operation as well as taking inverses, assume $a\in A$ and $g\in G\setminus A$. If $ag$ were in $A$, then $a^{-1}(ag) = g$ would be in $A$, which by assumption is not true. Hence $ag\notin A$.

We don't really gain much by not assuming that $A$ is a subgroup, since actually $A$ is a subgroup if and only if it is nonempty and closed under inverses and the operation. So this result also holds (vacuously) for the empty set.

Answer 2

Firstly, since $H$ is a subgroup, $g' \in H \Rightarrow g'^{-1}\in H$, and thence (contrapositive) $g'^{-1}\in \complement_GH \Rightarrow g'\in \complement_GH$. Now call your $g$ my $g'^{-1}$, and you get that $g \in \complement_GH \Rightarrow g^{-1}\in\complement_GH$.

Now, take $h\in H$; suppose $gh\in H$; we'd have:

$$(gh)^{-1}=h^{-1}g^{-1}\in H \Rightarrow h(h^{-1}g^{-1})=(hh^{-1})g^{-1}=g^{-1}\in H$$

contradiction. Then you're right, $gh\notin H$.

More generally, if we denote by $f$ group's operation, the following holds:

• $f(G \times G) \subseteq G$
• $f(H \times H) \subseteq H$
• $f(H \times \complement_GH) \subseteq \complement_GH$
• $f(\complement_GH \times H) \subseteq \complement_GH$
• $f(\complement_GH \times \complement_GH)=\begin{cases}\emptyset&\text{if }H=G\\H&\text{if }[G:H]=2\\G&\text{otherwise}\end{cases}$