Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 3 / 5E (Groups)

For a group action of $G$ on $X$, let $G_x=\{g:gx=x\}$ be the stabilizer subgroup of $x$, and $Gx$ its group orbit. The orbit-stabilizer theorem is the bijectionIt is well defined because elements in the same left coset differ by an element fixing $x$. Conversely, $gx=hx$ implies $h^{-1}g\in G_x$, so the cosets coincide. Surjectivity is the definition of the group orbit. For a finite group, counting the cosets givesTo obtain Cayley theorem, let $G$ act on its own underlying set by left multiplication. The map ${\lambda }_g:x\mapsto gx$ is a permutation, with inverse ${\lambda }_{g^{-1}}$, and ${\lambda }_g{\lambda }_h={\lambda }_{gh}$. If ${\lambda }_g$ is the identity permutation, evaluating at the identity element gives $g=e$. Thus $g\mapsto {\lambda }_g$ is an injective group homomorphism. When $|G|=n$, labeling this set by $n$ letters identifies its image with a subgroup of $S_n$. Every finite group embeds in the symmetric group on its own elements.

Choose $x_0\in X$ with stabilizer subgroup $H$. Transitivity means that every point is $g{x}_0$ for some $g$. Its stabilizer subgroup issince $a(g{x}_0)=g{x}_0$ exactly when $g^{-1}ag\in H$. The kernel of a group homomorphism arising from the group action is the set of elements fixing every point, henceThis is the subgroup core. A group homomorphism is injective precisely when its kernel is trivial. The desired equivalence is therefore exactly the equivalence between a trivial subgroup core and a faithful group action. No finiteness assumption is required in this part.

Every nontrivial subgroup of the quaternion group contains $-1$. Indeed, it either contains $-1$ directly, or contains one of $\pm i,\pm j,\pm k$, whose square is $-1$.

Suppose $Q_8$ acts on an $n$-point set with $n<8$. Every group orbit has size at most $n<8$. By the orbit-stabilizer theorem, each point's stabilizer subgroup has order $8/|Q_{8}x|>1$, so it contains $-1$. Consequently $-1$ fixes every point, and the group action is not faithful. An embedding in $S_n$ would give a faithful group action on $n$ letters, which is impossible. ThusThe left-regular group action on eight points is faithful by Cayley theorem, so the minimal faithful permutation degree of the quaternion group is exactly eight. The proof allows multiple group orbits; restricting only to transitive actions would not suffice.