Group Actions: Examples and Counterexamples

Dividing any integer by $2$ shows that the two residue classes are $0$ and $1$; they are distinct because $1-0$ is not a multiple of $2$, and $1+1=0$.

The rule $a\cdot \ast =\ast$ satisfies $0\cdot \ast =\ast$ and $(a+b)\cdot \ast =\ast =a\cdot (b\cdot \ast )$, so it is an action.

There is only one point of $X$, so the action is transitive.

The nonidentity class $1$ fixes $\ast$, and therefore fixes every point of $X$. Hence the action is not faithful.

The identities $e\cdot x=ex=x$ and $(gh)\cdot x=(gh)x=g(hx)=g\cdot (h\cdot x)$ verify the action laws.

Given $x,y\in G$, the element $g=y{x}^{-1}$ satisfies $g\cdot x=y$, so the action is transitive.

If $g\cdot x=x$, then $gx=x$ and right cancellation gives $g=e$; by [L2] the action is free, and [L3] gives $G_x=\{e\}$ for every $x$.

Evaluation satisfies $e(i)=i$ and $(\sigma \tau )(i)=\sigma (\tau (i))$, so [L4] and [L5] give an action of $S_3$ on $X$.

For any $i,j\in X$, a permutation carries $i$ to $j$, so the action is transitive. If a permutation fixes all three points, it is the identity function, so the action is faithful.

The nonidentity transposition $(23)$ fixes $1$, so $G_1$ is nontrivial and the action is not free by [L2] and [L3]. Thus the false claim fails.

The identity class satisfies $0\cdot (b,i)=(b,i)$, and $(a+c)\cdot (b,i)=(a+c+b,i)=a\cdot (c\cdot (b,i))$, so [L1] and [L4] give an action.

If $a\cdot (b,i)=(b,i)$, then $a+b=b$ in $\mathbb{Z}\mathrm{/}2$ and cancellation gives $a=0$; hence the action is free by [L2].

The second coordinate is unchanged by the action, so the two sets $A\times \{0\}$ and $A\times \{1\}$ are distinct orbits by [L3]. The action is not transitive, refuting the claim.

One has $0\cdot x=x$ and $(a+b)\cdot x=(a+b)+x=a+(b+x)=a\cdot (b\cdot x)$, so [L1] and [L3] give an action.

Given vertices $x,y$, the unique class $a=y-x$ satisfies $a\cdot x=y$, proving transitivity. If $a\cdot x=x$, cancellation gives $a=0$, so the action is free.

An element fixing every vertex fixes $0$, so it is $0$ by step 2.1; hence the action is faithful.

The set $H$ contains $0$, is closed under addition since $3+3=0$, and contains additive inverses; hence $H\le G$ by [L2], [L3], and [L4]. Its cosets are $H=\{0,3\}$, $1+H=\{1,4\}$, and $2+H=\{2,5\}$.

Since $G$ is abelian, every conjugate of $H$ is $H$, so ${\mathrm{Core}}_G(H)=H$.

By [L1], the coset action is transitive and has kernel $H$. Since $3\mathrm{\ne }0$ lies in the kernel, the action is not faithful.

The identity, the three transpositions, and the two $3$-cycles are six distinct permutations; by [L3] and [L5], they exhaust $S_3$.

Conjugation relabels cycle entries: every transposition is conjugate to every other, and the two $3$-cycles are conjugate. Cycle type is preserved by conjugation, so the three displayed sets are exactly the conjugacy classes.

Their cardinalities are $1$, $2$, and $3$, so [L1] gives $6=1+2+3$.

An element is central exactly when its conjugacy class is a singleton. By [L2], the only such element of $S_3$ is $e$, so $Z(S_3)=\{e\}$.

The group $S_3$ is nontrivial and finite but has trivial center. Thus the false claim fails, and [L1] shows precisely why the $p$-group hypothesis cannot be omitted.

Pointwise calculation gives $r^4=e$, $s^2=e$, and $sr=r^{-1}s$.

The relations reduce every word in $r,s$ to one of $e,r,r^2,r^3,s,rs,r^{2}s,r^{3}s$ and show this set is closed under products and inverses. These eight permutations are distinct, so [L3] and [L4] make them a subgroup $D$ of order $8$; also $sr\mathrm{\ne }rs$, so $D$ is nonabelian.

Using $sr=r^{-1}s$, conjugation by $r$ and $s$ gives the five displayed conjugacy classes: $e$ and $r^2$ are central, $r$ is conjugate to $r^3$, and the reflections split into the two displayed pairs.

Their sizes give $8=1+1+2+2+2=2+2+2+2$, and [L1] identifies the two singleton classes with the center $\{e,r^2\}$.

By [L2], $D$ is nonabelian and has order $8=2^3$.

Since $2$ is prime, $D$ refutes the false claim and shows that the exponent $2$ in [L1] cannot be replaced by $3$.

Each transposition generates a two-element subgroup, and these are distinct. Every element of order $2$ in $S_3$ is a transposition, so these are all the subgroups of order $2$.

Conjugation relabels the two moved points, so the three subgroups in step 1.1 are conjugate.

By [L1], $[S_3:N_{S_3}(H)]=3$. Since $\mathrm{\mid }{S}_3\mathrm{\mid }=6$ by [L5], the normalizer has order $2$; it contains $H$, so it equals $H$. The same holds for each conjugate subgroup.

Map $0$ to the identity permutation and $1$ to $(12)$. Since $(12{)}^2=e$, [L2] and [L3] give an action of $P$ on $X$.

Its orbit partition is $\{1,2\},\{3\},\{4\},\{5\}$, and the global fixed set is $\{3,4,5\}$.

Hence $\mathrm{\mid }X\mathrm{\mid }=5$, $\mathrm{\mid }{X}^P\mathrm{\mid }=3$, and $5-3=2$ is divisible by $2$, verifying [L1] by [L4].

For each $i\in X$, some transposition moves $i$, so no point is fixed by every element of $S_3$ and $X^{S_3}=\mathrm{\varnothing }$. Thus $\mathrm{\mid }X\mathrm{\mid }=3$ and $\mathrm{\mid }{X}^{S_3}\mathrm{\mid }=0$.

The difference $3-0=3$ is not divisible by $2$, so [L5] shows that the congruence fails. By [L4], $\mathrm{\mid }{S}_3\mathrm{\mid }=6$ is not a power of $2$, so this does not contradict [L1] and isolates its $p$-group hypothesis.

Direct multiplication shows that $A$ consists of the identity, the eight $3$-cycles on $\{1,2,3,4\}$, and the three double transpositions $(12)(34)$, $(13)(24)$, $(14)(23)$. This set is closed under multiplication by $a^{\pm 1}$ and $b=b^{-1}$, and every listed element is a word in $a,b$, so [L3] and [L4] give $\mathrm{\mid }A\mathrm{\mid }=12$.

Suppose, for contradiction, that $H\le A$ has order $6$. Then [L5] gives $[A:H]=2$, so $H⊴A$ by [L2]. Since $3\mid \mathrm{\mid }H\mathrm{\mid }$, [L1] gives an element $c\in H$ of order $3$, necessarily one of the eight $3$-cycles in step 1.1.

The subgroup $H$ contains $c^{-1}$. Normality and [L6] make it contain every $A$-conjugate of both $c$ and $c^{-1}$. Conjugating by the displayed generators gives two disjoint four-element classes, together all eight $3$-cycles, so $H$ would contain at least eight elements, contradicting $\mathrm{\mid }H\mathrm{\mid }=6$.

Thus $6$ divides $\mathrm{\mid }A\mathrm{\mid }=12$ but is not the order of a subgroup of $A$. The arbitrary-divisor claim is false, whereas [L1] remains valid for prime divisors.

Define $(g\cdot c)(v)=c(g^{-1}v)$. Inverse precomposition gives $e\cdot c=c$ and $(gh)\cdot c=g\cdot (h\cdot c)$, so [L2] and [L3] give an action on the $16$ colourings counted by [L4].

A colouring fixed by a symmetry is constant on each cycle of that symmetry. Thus the identity fixes $16$ colourings; the two quarter-turns fix $2$ each; the half-turn fixes $4$; the two reflections through opposite vertices fix $8$ each; and the two reflections through opposite edges fix $4$ each.

The fixed-point sum is $16+2+2+4+8+8+4+4=48$. By [L1] and $\mathrm{\mid }D\mathrm{\mid }=8$, one has $48=8\mathrm{\mid }C\mathrm{/}D\mathrm{\mid }$, so $\mathrm{\mid }C\mathrm{/}D\mathrm{\mid }=6$.

The six orbits can also be distinguished by the number of black vertices, with the two-black case split into adjacent and opposite pairs, confirming the count.

Let $(a\cdot c)(i)=c(i-a)$. Addition in [L3] verifies the action laws [L2], and [L4] and [L5] give four rotations acting on $16$ words.

Rotation by $0$ fixes all $16$ words; rotations by $1$ and $3$ each force all positions equal and fix $2$ words; rotation by $2$ permits one colour on each opposite pair and fixes $4$ words.

The fixed-point sum is $16+2+4+2=24$, so [L1] gives $24=4\mathrm{\mid }C\mathrm{/}R\mathrm{\mid }$ and hence $\mathrm{\mid }C\mathrm{/}R\mathrm{\mid }=6$.

By [L2], the singleton is one orbit, so $\mathrm{\mid }X\mathrm{/}G\mathrm{\mid }=1$.

Here $\mathrm{\mid }X\mathrm{\mid }=1$ and $\mathrm{\mid }G\mathrm{\mid }=2$, so there is no natural number $q$ with $\mathrm{\mid }X\mathrm{\mid }=\mathrm{\mid }G\mathrm{\mid }q$; in particular the orbit count is not obtained by dividing $\mathrm{\mid }X\mathrm{\mid }$ by $\mathrm{\mid }G\mathrm{\mid }$.

Both elements of $G$ fix the unique point, so [L1] correctly gives $2\cdot 1=1+1$. This verifies the orbit-counting identity while refuting the naive division rule.