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A diagonal of a polygon joins two vertices of a polygon that are not endpoints of the same side of a polygon. A square has two diagonals of a polygon, forming a two-element group orbit under its dihedral group of symmetries.

For a group action of any group $G$ on a set $X$, the cardinality of the orbit of a group action through $x$ is the index of a subgroup of its stabilizer subgroup:In particular, if $G$ is a finite group,For a fixed finite group, a larger stabilizer subgroup means a smaller group orbit: more transformations repeat the same image of $x$. The set $X$ itself need not be finite.

The underlying bijection uses left cosets, without requiring $G_x$ to be a normal subgroup:To check it briefly, replacing $g$ by $gh$ with $h\in G_x$ preserves $g\cdot x$, so $\Phi$ is well-defined. Every member of the group orbit has this form. Finally, $g\cdot x=k\cdot x$ implies $k^{-1}g\in G_x$, hence $g{G}_x=k{G}_x$, proving the bijection. This is the construction in MIT's group-action notes, lemma 23.5. For infinite groups, the bijection and the cardinality statement $|Gx|=[G:G_x]$ still hold; division of infinite cardinalities does not supply the finite formula.

For the eight-element dihedral group $D_4$ acting on the square described under group action, moving a marked vertex of a polygon gives $8=4\cdot 2$. Moving a marked side of a polygon also gives $8=4\cdot 2$; moving one unoriented diagonal of a polygon gives $8=2\cdot 4$; moving the centroid gives $8=1\cdot 8$. These are four different group actions, each with one group orbit. In the first group action, the four left cosets of $H=G_A=\{e,s\}$ areThey correspond respectively to $A,B,C,D$. Counting all eight transformations would count each position of the marked vertex of a polygon twice.

For a second computed example, let the symmetric group $S_3$ act on the Cartesian product $X=\{1,2,3{\}}^2$ by $\sigma \cdot (i,j)=(\sigma (i),\sigma (j))$. There are two orbits of a group action:andThe first stabilizer subgroup can exchange $2$ and $3$ while fixing $1$; the second must fix both $1$ and $2$, so also fixes $3$. The orbit-stabilizer theorem gives $6=3\cdot 2$ and $6=6\cdot 1$, and the two group orbit sizes account for all nine ordered pairs. The general reason for exactly these two group orbits is equality-pattern orbits of ordered pairs.

Take the electric potential to vanish at infinity. The distance from the centre to each point charge is half the diagonal of a polygon of the square, namely $a/\sqrt{2}$. The superposition principle for the electric potential therefore gives, for four charges $+q$,Strictly, specifying only the magnitude of each electric charge leaves its sign undetermined. For signed charges $q_j$, the general answer is $V=\sqrt{2}{\sum }_{j=1}^4{q}_j/(4\pi {ϵ}_{0}a)$. For example, two positive and two negative charges of equal magnitude give zero electric potential at the centre. The boxed result uses the usual interpretation that all four charges have the same positive sign.

For a group action of a group $G$ on a set $X$, the stabilizer subgroup of $x\in X$ consists of the transformations that leave $x$ fixed:It is a subgroup: the identity element fixes $x$, products of transformations fixing $x$ fix $x$, and the inverse element of a transformation fixing $x$ also fixes $x$. A stabilizer subgroup is a subset of $G$, whereas the orbit of a group action is a subset of $X$. See Ackerman's algebra notes, definition 0.4.1.2.

For the square group actions defined under group action, direct calculation givesThe reflection $s$ fixes $A$ and $C$ individually. The reflection $rs$ preserves the unoriented side of a polygon $AB$ by exchanging its two vertices of a polygon. The half-turn rotation $r^2$ exchanges the two ends of the unoriented diagonal of a polygon $AC$ while preserving that diagonal of a polygon. Thus fixing a side of a polygon or diagonal of a polygon as an object does not mean fixing every point on it; the latter condition uses a pointwise stabilizer. The four stabilizer subgroups have group orders $2,2,4,8$, matching group orbit sizes $4,4,2,1$ through the orbit-stabilizer theorem.

A stabilizer subgroup records the transformations still available after imposing a constraint. Fixing a vertex of a polygon of the square leaves two choices; fixing its centroid leaves all eight. In a faithful group action, a stabilizer subgroup can still contain a nonidentity transformation, as $s\in (D_4{)}_A$ shows: a faithful group action requires that only the identity element fix every point of $X$ simultaneously, not that every individual stabilizer subgroup be trivial. The two-point cyclic group example under group action has both stabilizer subgroups equal to the group kernel $\{0,2\}$ of its associated group homomorphism.