Question is to prove that :
$Q_8$ is isomorphic to a subgroup of $S_8$, but not isomorphic to a subgroup of $S_n$ for $n\leq 7$.
I see that $Q_8$ is isomorphic to subgroup of $S_8$ by left multiplication action.
Hint given was to prove that stabilizer of any point contains $\{\pm 1\}$.
To prove Cayley's theorem, stating any group is isomorphic to a subgroup of $S_n$ we take action of given group on a set $A$ having same cardinality.
with that motivation I want to check if there is a Isomorphism then there is a map from $G\times A \rightarrow A$. i.e., $G$ gives a permutation group $S_A$.
I tried in same manner. Suppose $Q_8$ is isomorphic to subgroup of $S_n$ with $n\leq 7$ then it should come from a group action of $Q_8$ on a set of cardinality atmost 7.
Suppose $Q_8$ acts on a set $A$ with possible cardinality at most 7.
$stab(a)=\{g\in Q_8 : g.a=g \forall a\in A \}$
$cl(a)=\{g.a : g\in Q_8\}$
I know number of elements in class of $a$ equals to index of stabilizer.
As $cl(a)=\{g.a : g\in Q_8\}\in A$ i.e., $cl(a)\subseteq A$ and as $|A|\leq 7$ i see that $|cl(a)|\leq 7$.
But, $|cl(a)|=|Q_8:stab(a)|$ for any element $a\in A$.
So, $|Q_8:stab(a)|=|cl(a)|\leq 7$ for all $a\in A$.
So, $stab(a)$ should be non trivial subgroup of $Q_8$ if not then $|Q_8:stab(a)|=8$
non trivial subgroup (proper) of $Q_8$ contains $\{\pm1\}$.
So, In the worst case, $\{\pm 1\}\subseteq stab(a)$ for all $a\in A$.
As $Ker(\eta)=\cap_{a\in A}stab(a)$ (where $\eta$ is the action of $Q_8$ on $A$)
we see that $\{\pm 1\}\subseteq Ket(\eta)$ which means that $Ker(\eta)$ is non trivial.
thus there is no isomorphism (coming from $\eta$) between $Q_8$ and any subgroup of $S_7$.
I would be thankful if someone can check whether my approach is correct or if there is any other simple possible way.
P.S : Usually what i do to see whether two groups are isomorphic or not is to check for cardinality, abelian property, no of elements with same order and so on. But I was having no idea when i fail in all these ways. With this Group actions i could see possibility for getting a precise conclusion on Isomorphisms.I would like to Thank Mr. Jyrki Lahtonen (a user of Math.SE) who made me to get used to Group actions.
P.S $2$: If any thing is wrong in my idea, it is entirely my fault, and if anything is correct in this whole credit should go to Mr. Jyrki Lahtonen
Best Answer
The proof is correct, but one can generalize and shorten it as follows:
Let $G$ be a finite group and assume that the intersection of all non-trivial subgroups of $G$ is non-trivial, i.e. contains some $g \neq 1$. If $G$ acts on a set $A$ with $|A|<|G|$, then for every $a \in A$ we have $|G:G_a|=|Ga| \leq |A| < |G|$. Therefore, $G_a$ is non-trivial, and $g \in \cap_{a \in A} G_a = \ker(G \to \mathrm{Sym}(A))$. Hence, $G$ doesn't embed into $\mathrm{Sym}(A)$.
For the Quaternion group we can take $g=-1$.