Exercise 4.2.7 ($Q_8$ is not isomorphic to a subgroup of $S_n$ for $n\leq 7$)

Proof. Let $G=Q_8$ be the quaternion group of order $8$.

(a)

By Exercise 4, the left regular representation of $Q_8$ affords the isomorphism $$\begin{array}{llll}\hfill Q_8& \simeq ⟨(1324)(5768),(1526)(3847)⟩& \hfill & \\ \hfill & & \hfill \end{array}$$

which shows that $Q_8$ is isomorphic to a subgroup of $S_8$.

(b)

We suppose that $Q_8$ acts on a set $A$, where $|A|\le 7$ (this is possible: for instance, $Q_8$ acts on the left cosets of $⟨-1⟩$).

Let $a\in A$, and assume for the sake of contradiction that $-1\notin G_a$, so that

$$b=(-1)\cdot a\ne a.$$

Then $(-1)\cdot b=(-1)\cdot ((-1)\cdot a)={(-1)}^2\cdot a=1\cdot a=a.$

Consider the set $B=\{a,b\}$. We show that $B$ is a block, as defined in Exercise 4.1.7, i.e. for all $g\in D_8$,

$$g\cdot B=B\text{or}B\cap g\cdot B=\varnothing .$$

• If $g=1$ then $1\cdot B=B$, and if $g=-1$, then

  $$(-1)\cdot B=\{(-1)\cdot a,(-1)\cdot b\}=\{b,a\}=B.$$

• If $g=i$, we show that $B\cap \mathit{𝑖𝐵}=\{a,b\}\cap \{i\cdot a,i\cdot b\}=\varnothing$.

  • If $i\cdot a=a$, then

    $$b=(-1)\cdot a=i\cdot (i\cdot a)=i\cdot a=a;$$

    This is impossible, since $b\ne a$.

  • If $i\cdot a=b$, then $i\cdot b=i\cdot (i\cdot a)=(-1)\cdot a=b$, thus

    $$a=(-1)\cdot b=i\cdot (i\cdot b)=i\cdot b=b,$$

    which is also impossible, since $b\ne a$.

    Therefore $i\cdot a\notin \{a,b\}$, and similarly $i\cdot b\notin \{a,b\}$, so

    $$\begin{array}{lll}\hfill B\cap i\cdot B=\varnothing .& & \hfill \text{(1)}\end{array}$$

• If $g=j$, or $g=k$, we obtain similarly (because there is an automorphism of $Q_8$ which maps $i$ on $j$ (or $i$ on $k$))

  $$\begin{array}{lll}\hfill B\cap j\cdot B=\varnothing ,B\cap k\cdot B=\varnothing .& & \hfill \text{(2)}\end{array}$$

  Therefore, for every $g\in G$ and for every $h\in G$,

  $$g\cdot B=h\cdot B\text{or}g\cdot B\cap h\cdot B=\varnothing :$$

  if $g\cdot B\ne h\cdot B$, then $(h^{-1}g)\cdot B\ne B$, thus $(h^{-1}g)\cdot B\cap B=\varnothing$, so $g\cdot B\cap h\cdot B=\varnothing$.

  Note also that $i\cdot B\ne j\cdot B$, otherwise $\mathit{𝑖𝐵}=\mathit{𝑗𝐵}$, thus $k\cdot B=(j^{-1}i)\cdot B=B$, in contradiction with (2). Similarly $j\cdot B\ne k\cdot B$ and $i\cdot B\ne k\cdot B$.

  Let ${𝒪}_a=\{g\cdot a\mid g\in Q_8\}\subseteq A$, so that $Q_8$ acts transitively on ${𝒪}_a$. Then the distinct sets

  $$B,i\cdot B,j\cdot Bk\cdot B$$

  form a partition of ${𝒪}_a$ (since $B$ is a block, this is also a consequence of Exercise 4.1.7(b)).

  Therefore

  $$|A|\ge |{𝒪}_a|=|B|+|i\cdot B|+|j\cdot B|+|k\cdot B|=8,$$

  in contradiction with the hypothesis $|A|\le 7$. This contradiction shows that $-1\in G_a$, so

  $$⟨-1⟩\le G_a.$$

  Reasoning by contradiction, assume that $Q_8\simeq H\le S_n(n\le 7)$. Let $f:Q_8\to H$ be an isomorphism. Then $Q_8$ acts on $A=[[1,n]]$, the action being defined by

  $$g\cdot i=f(g)(i),(i\in [[1,n]],g\in Q_8).$$

  Since the action of $H$ is faithful, the action of $Q_8$ is faithful: if $g\cdot i=i$ for all $i\in A$, then $f(g)(i))=i$ for all $i$, thus $f(g)=1_A$, where $f$ is an isomorphism, so $g=1$.

  Let $a\in A$ (for instance $a=1$), and consider the restriction of this action to ${𝒪}_a$. Then $Q_8$ acts faithfully and transitively on ${𝒪}_a$.

  By the first part of the proof, $(-1)\cdot a=a$. If $c$ is any element of ${𝒪}_a$, there is some $g\in Q_8$ such that $g\cdot a=c$. Therefore

  $$\begin{array}{llllll}\hfill (-1)\cdot c& =(-1)\cdot (g\cdot a)& \hfill & & \hfill & \\ \hfill & =((-1)g)\cdot a& \hfill & & \hfill & \\ \hfill & =(g(-1))\cdot a& \hfill (-1\in Z(Q_8))& & \hfill & & \hfill \\ \hfill & =g\cdot ((-1)\cdot a)& \hfill & & \hfill & \\ \hfill & =g\cdot a& \hfill & & \hfill & \\ \hfill & =c.& \hfill & & \hfill & \end{array}$$

  Since $(-1)\cdot c=c$ for all $c\in {𝒪}_a$, this shows that $-1$ is in the kernel of the action of $Q_8$ on ${𝒪}_a$, so the action of $Q_8$ on ${𝒪}_a$ is not faithful.

  This is a contradiction, which shows that $Q_8$ is not isomorphic to a subgroup of $S_n$ for any $n\le 7$.