0
$\begingroup$

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 $n\leq 7$.

$\endgroup$
  • 1
    $\begingroup$ Have you learned Sylow's theorems? If so, and you can find one subgroup of order $8$... $\endgroup$ – pjs36 Nov 12 '17 at 23:22
  • $\begingroup$ Let $f(x) = \prod_{j=1}^n (x-\alpha_j) \in K[x]$ and $L = K(\alpha_1,\ldots,\alpha_n)$ its splitting field. Then the Galois group of $f $ is the group of those permutations of the $\alpha_j$ corresponding to automorphisms $\in Aut(L/K)$. $\endgroup$ – reuns Nov 12 '17 at 23:24
  • $\begingroup$ @reuns: Sorry. Brain fart. Thanks for putting me right. $\endgroup$ – Rob Arthan Nov 13 '17 at 0:02
2
$\begingroup$

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 $$ \langle \pm 1 \rangle \subset \bigcap_{i \in \{1...n\}} Q_{8_i} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.