$Q_8$ is isomorphic to a subgroup of $S_8$, but not isomorphic to a subgroup of $S_n$ for $n\leq 7$.

Question

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 at most 7.

Suppose $Q_8$ acts on a set $A$ with possible cardinality at most 7. For $a \in A,$ let

$\textrm{Stab}(a)=\{g\in Q_8 : g \cdot a = a\}$

$\textrm{cl}(a)=\{g \cdot a : g\in Q_8\}$

denote the stabilizer of $a$ in $Q_8$ and the class (or orbit) of $a,$ respectively. I know the number of elements in a class of $a$ equals to the index of the stabilizer of $a$ in $Q_8.$

By the definition of class of $a$, $\textrm{cl}(a)\subseteq A$ and as $|A|\leq 7,$ I see that $|\textrm{cl}(a)|\leq 7$.

But, $|\textrm{cl}(a)|=|Q_8: \textrm{Stab}(a)|$ for any element $a\in A$.

So, $|Q_8:\textrm{Stab}(a)|=|\textrm{cl}(a)|\leq 7$ for all $a\in A$.

So, $\textrm{Stab}(a)$ should be non-trivial subgroup of $Q_8$ (if not then $|Q_8:\textrm{Stab}(a)|=8.$

A proper non-trivial subgroup of $Q_8$ contains $\{\pm1\}$.

So, in the worst case, $\{\pm 1\}\subseteq \textrm{Stab}(a)$ for all $a\in A$.

As $\ker(\eta)=\cap_{a\in A}\textrm{Stab}(a)$ (where $\eta$ is the action of $Q_8$ on $A$)

We see that $\{\pm 1\}\subseteq \ker(\eta)$ 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

Answer 1

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 $g=-1$.

Answer 2

@Martin Brandenburg - the class of finite groups $G$ whose intersection of all non-trivial subgroups is non-trivial, can be specified further as follows. Put $$\mu(G)=\bigcap\{H : 1 \lt H \leq G\}$$ First of all, if we assume $\mu(G) \neq 1$, then $G$ must be a $p$-group for some prime $p$. If not then pick two different primes $p$ and $q$ dividing the order of $G$. Cauchy's Theorem guarantees the existence of subgroups $H \cong C_p$ and $K \cong C_q$, but then $\mu(G) \subseteq H \cap K=1$, a contradiction. In addition, $\mu(G) \unlhd G$, since each conjugate $\mu(G)^g$ is a subgroup for any $g \in G$, so $\mu(G) \subseteq \mu(G)^g$ and normality follows. It is also clear that $\mu(G)$ is the unique minimal normal subgroup of $G$ and for that matter, also a normal minimal subgroup. ($G$ is said to be monolithic).

In a $p$-group each central element of order $p$ generates a normal minimal subgroup. So the $p$-groups with a unique normal minimal subgroup are exactly the $p$-groups with cyclic center. Taking this a step further, $p$-groups with a unique subgroup of order $p$ (hence unique normal subgroup) must be either cyclic or generalized quaternion (cf. Theorem $5.46$ in Rotman's Introduction to the theory of groups 4th edition, or see Keith Conrad, Theorem $4.9$). So your remark generalizes to this class of groups.