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The quaternion group is not a subgroup of Symmetric group for any n less than 8


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.2 Exercise 4.2.7

Solution:

(1) $Q_8$ is a subgroup of $S_8$ via the left regular representation.

(2) Now suppose that $Q_8$ acts on a set $A$ of order at most $7$. For any element $a \in A$, we have $$[Q_8 : \mathsf{stab}(a)] = |Q_8 \cdot a| \leq 7.$$ By Lagrange’s Theorem, then, $|\mathsf{stab}(a)| \in \{2,4,8\}$. That is, $\mathsf{stab}(a)$ is nontrivial for all $a$.

We know that $\langle -1 \rangle$ is the unique $\leq$-minimal nontrivial subgroup of $Q_8$, so that $\langle -1 \rangle \leq \mathsf{stab}(a)$ for all $a \in A$. Thus if $K$ is the kernel of this action, $\langle -1 \rangle \leq K$ so that the action is not faithful.

Now if $Q_8$ is isomorphic to a subgroup of $S_7$, then the injection $Q_8 \rightarrow S_7$ induces a faithful action of $Q_8$ on the set $\{1,2,\ldots,7\}$. However, we showed that no such action can be faithful. Thus $Q_8$ is not isomorphic to a subgroup of $S_7$, and since $S_k \leq S_7$ for $1 \leq k \leq 7$, $Q_8$ is not isomorphic to a subgroup of $S_k$ for all $1 \leq k \leq 7$.


Linearity

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