Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics I / 7

A left group action of $G$ on a set $S$ is a function $G\times S\to S$, written $(g,s)\mapsto g\cdot s$, satisfyingfor all $g,h\in G$, $s\in S$. The orbit of a group action through $s$ and the stabilizer subgroup of $s$ are, respectively,The identity element belongs to $G_s$. If $g,h\in G_s$, then $(gh)\cdot s=g\cdot (h\cdot s)=g\cdot s=s$, so $gh\in G_s$. If $g\in G_s$, applying $g^{-1}$ to $g\cdot s=s$ gives $s=g^{-1}\cdot s$, so $g^{-1}\in G_s$. These facts prove that the stabilizer subgroup is a subgroup of $G$.

The orbit-stabilizer theorem identifies the orbit of a group action of $s$ with the set of left cosets of $G_s$. Explicitly, there is a bijectionHere $G/G_s$ denotes a set of cosets; it is not asserted to be a quotient group. To prove the identification, observe thatThis equivalence proves both well-definedness and injectivity. Every element of the orbit of a group action has the form $g\cdot s$, proving surjectivity.

In particular, for a finite group, the cardinalities satisfyThe last equality follows from Lagrange's theorem applied to the stabilizer subgroup.

The orbits of a group action form a partition of a set $S$: every element lies in its own orbit of a group action, and two intersecting orbits of a group action coincide because each point can be moved back and then forward using inverse elements. By the orbit-stabilizer theorem, each orbit of a group action has cardinality dividing $8$, so its size is $1,2,4$ or $8$.

If every orbit of a group action had size $2,4$ or $8$, their sizes would add to an even number, contrary to $|S|=17$. Therefore there is a one-element orbit of a group action, say $\{s^{\prime }\}$. Every $g\in G$ sends $s^{\prime }$ into this same singleton, soThus $s^{\prime }$ is a fixed point of a group action. In fact the number of fixed points of a group action is odd, since all other orbits of a group action have even cardinality.

For a free group action, every stabilizer subgroup is $\{e\}$. By the orbit-stabilizer theorem, every orbit of a group action has cardinality $|G|$. Hence the two free group actions have the same numberof orbits of a group action. Choose one representative $s_i$ from each $\cdot$ orbit of a group action and one representative $t_i$ from each $\ast$ orbit of a group action, and match them using $1\le i\le r$.

Every $s\in S$ has a unique expression $s=g\cdot s_i$. The orbit of a group action determines $i$, and the free group action property determines $g$: if $g\cdot s_i=h\cdot s_i$, then $(h^{-1}g)\cdot s_i=s_i$, forcing $g=h$. Therefore the ruledefines a function $f:S\to S$ without ambiguity. For $h\in G$,which proves the required identity for every $s$.

Indeed, $f$ is a bijection: its inverse is $g\ast t_i\mapsto g\cdot s_i$, well defined because the second is also a free group action. If $S$ is empty, the unique empty function already satisfies the identity. Thus the result holds in every case, and proves the stronger statement that free finite actions with equal cardinality are isomorphic.