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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

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  • $\begingroup$ Thanks. Appreciated, but no need for credits like that within a question body. <Blush> (comments are fine :-)</Blush>. We all do our part, and so do you. $\endgroup$ – Jyrki Lahtonen Sep 2 '13 at 9:46
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    $\begingroup$ If that credits bothers you, Sorry. But i am not going to take that credits back :) $\endgroup$ – user87543 Sep 2 '13 at 9:49
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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 $g=-1$.

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  • $\begingroup$ PS: $Ga$ denotes the orbit of $a$ and $G_a$ denotes the stabilizer of $a$. $\endgroup$ – Martin Brandenburg Sep 2 '13 at 10:38
  • $\begingroup$ I wish i could write like this. More precisely in simple words :) $\endgroup$ – user87543 Sep 2 '13 at 10:56

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