Let $G$ be an arbitrary group.Let $S$ and $H$ be subgroups of $G$ which are not equal with $G$. Show that $G\setminus H$ is not subset of $S$. Let $G$ be an arbitrary group.
Let $S$ and $H$ be subgroups of $G$ which are not equal with $G$. Show that $G\setminus H$ is not subset of $S$.
$S$ subgroup of $G$, then for all $s_1, s_2 \in S, s_1s_2^{-1} \in S$ and $H$ subgroup of $G$, then for all $h_1,h_2 \in H, h_1h_2^{-1} \in H$. Let $G = \{e_G,g,g^2,\dots, g^{n-1}\}$.
What next?
Any idea? Thanks in advance.
 A: So you need to prove that $G\ne S\cup H$. Suppose that $G=S\cup H$ and $S\ne G\ne H$. If $H\subseteq S$ or $S\subseteq H$, then $G=H\cup S$ is either $S$ or $H$, a contradiction.
Hence  $S\not\subseteq H$ and $H\not\subseteq S$. Let $s\in S\setminus H$ and $h\in H\setminus S$. Since $sh\in G$ it is either in $S$ or in $H$. In the first case $h=s^{-1}(sh)\in S$, a contradiction. In the second case $s=(sh)h^{-1}\in H$, a contradiction.
Hence $S\cup H\ne G$.
A: Hint:  It is well-known that the union of two subgroups is again a subgroup if and only if the subgroups are nested, that is $S\le H $ (wlog).
Combine this with the fact that if $G-H\subset S$, then $G=H\cup S$.
Finally if $S$ and $H$ are nested and neither is all of $G$, we are done.
A: See also the answer of @JCCA: a group cannot be written as the union of two of its proper (:= being not equal to the whole group) subgroups. Hence it follows:

**Proposition** Let $H$ be a proper subgroup of a group $G$. Then $\langle G \setminus H \rangle=G$.
**Proof** Obviously $G=H \cup G \setminus H \subseteq H \cup \langle G \setminus H \rangle$. Since $H$ is proper the proposition follows.

So in your case, if $G \setminus H \subseteq S$. then, since $S$ is a subgroup, also $\langle G \setminus H \rangle \subseteq S$, a contradiction to $S$ being proper.
