Show that $Q_8$ is not isomorphic to a subgroup of $S_4$ , Conclude that $Q_8$ is not the Galois group of the splitting field of a degree $4$ Let $Q_8$ be the quaternion group $\{±1, ±i, ±j, ±k\}$, where multiplication
is determined by the relations $i^2=j^2=-1$ and $ij = k = -ji$. 
Show that $Q_8$ is not isomorphic to a subgroup of $S_4$ , Conclude that
$Q_8$ is not the Galois group of the splitting field of a degree $4$ polynomial
over a field.
What are the subgroups of $ S_4 $? And those who have order $8$? I thought about applying the fundamental theorem here and saying that if $Q_8$ was isomorphic to a subgroup of $S_4$ then his fixed field would correspond to the group of Galois of $Q_8$, but I do not know what else to do, could someone help me please?
Here is something interesting that I found: $Q_8$ is isomorphic to a subgroup of $S_8$ but not to asubgroup of $S_n$ for $n\leq 7$.
 A: We can prove that $Q_8$ is not a subgroup of $S_n$ for $n\leq 7$ 
We use three basic facts:
$1.)$ Any non trivial subgroup of $Q_8$ contains $\langle \pm 1 \rangle$
$2.)$ The orbit stabilizer theorem.
$3.)$ If a group $G$ acts on a set $S$, the kernel of the action is $\bigcap_{s \in S} G_s$ where $G_s$ is the stabilizer of the element $s$
Proof: Suppose $Q_8$ was isomorphic to a subgroup of $S_n$ $n \leq 7$.  Then $Q_8$ acts on a set of $n$ elements faithfully.   
Then for any $i \in \{1,2...n\}$ with $n \leq 7$ , the orbit stabilizer theorem tells us that $(Q_8:Q_{8_i}) \neq 8$ (i.e we can't have an orbit containing $8$ elements when we're only acting on $7$ or less). So all the stabilizer subgroups are non-trivial subgroups since their indexes are smaller than $8$.  Since all non trivial subgroups of $Q_8$ contain $\langle \pm 1 \rangle$ any stabilizer subgroup contains $\langle \pm 1 \rangle$.  
But this contradicts that $G$ acts faithfully on $\{1,...n\}$ as the kernel of the action should be $\langle 1 \rangle = \bigcap Q_{8_i}$.  But $$ \langle \pm 1 \rangle \subset \bigcap_{i \in \{1...n\}} Q_{8_i} $$
