## Tag Archives: group action

### Every ring action induces a group action

Let $R$ be a ring with 1 and let $M$ be a left unital $R$-module. Prove that the restriction of the module operator $\cdot$ to $R^\times \times M$ is a group action operator on the group of units in $R$.

Recall that $R^\times$ is indeed a group; certainly the restriction of $\cdot$ to $R^\times \times M$ is a mapping $R^\times \times M \rightarrow M$. Now for all $m \in M$, we have $1 \cdot m = m$, and if $u,v \in R^\times$, then $(uv) \cdot m = u \cdot (v \cdot m)$. So $R^\times$ acts on $M$ by the restriction of our module operator $\cdot$.

### The Frattini subgroup is characteristic

Let $G$ be a group and denote by $\Phi(G)$ the Frattini subgroup of $G$. Prove that $\Phi(G) \leq G$ is characteristic.

We begin with some lemmas. Let $\mathcal{S}(G)$ denote the set of subgroups of $G$; recall that $\mathsf{Aut}(G)$ acts on $\mathcal{S}(G)$ by application: $\alpha \cdot H = \alpha[H]$.

Lemma 1: Let $G$ be a group. If $\mathcal{O}$ is an orbit of the action of $\mathsf{Aut}(G)$ on $\mathcal{S}(G)$, then $\bigcap \mathcal{O}$ is characteristic in $G$. Proof: Let $\alpha$ be an automorphism of $G$. Now $\alpha[\bigcap_{H \in \mathcal{O}} H]$ $= \bigcap_{H \in \mathcal{O}} \alpha[H]$ $= \bigcap_{H \in \mathcal{O}} H$, since $\alpha$ permutes the elements of $\mathcal{O}$. $\square$

Lemma 2: The set $\mathcal{M} \subseteq \mathcal{S}(G)$ of all maximal subgroups of $G$ is a union of orbits under the action of $\mathsf{Aut}(G)$. Proof: It suffices to show that if $M$ and $H$ are maximal and nonmaximal subgroups of $G$, respectively, then no automorphism carries $M$ to $H$. Suppose to the contrary that $\alpha$ is such an automorphism and that $H < K < G$. Then $G > \alpha^{-1}[K] > M$, a contradiction. $\square$

Now to the main result. We see that $\Phi(G) = \bigcap \mathcal{M}$ is an intersection of characteristic subgroups, hence characteristic.

### Sym(n) acts on the n-fold direct product of a group by permuting the factors

Let $G_1 = G_2 = \cdots = G_n$ and let $G = \times_{i=1}^n G_i$. Under the notation of the previous exercise, prove that $\varphi_\pi \in \mathsf{Aut}(G)$. Show also that the map $\pi \mapsto \varphi_\pi$ is an injective homomorphism $S_n \rightarrow \mathsf{Aut}(G)$.

(In particular, show that $\varphi_{\pi_1} \circ \varphi_{\pi_2} = \varphi_{\pi_1\pi_2}$. It is at this point that the $\pi^{-1}$ in the definition of $\varphi_\pi$ are needed. The underlying reason for this that if $e_i$ is the $n$-tuple with 1 in position $i$ and 0 elsewhere, then $S_n$ acts on $\{e_1,e_2,\ldots,e_n\}$ on the left by $\pi \cdot e_i = e_{\pi(i)}$. If the $n$-tuple $(g_1,g_2,\ldots,g_n)$ is represented by the formal sum $g_1e_1 + g_2e_2 + \cdots + g_ne_n$, then this left group action on the $e_i$ extends to a left group action on the set of all formal sums by $\pi \cdot (\sum_{i=1}^n g_ie_i) = \sum_{i=1}^n g_i e_{\pi(i)}$. Note that the coefficient of $e_i$ on the right hand side is $g_{\pi^{-1}(i)}$, thus the right hand side may be rewritten as $\sum_{i=1}^n g_{\pi^{-1}(i)} e_i$, which is precisely the formal sum representation of $\varphi_\pi((g_1,\ldots,g_n))$. In other words, any permutation of the “position vectors” $e_i$ (which fixes their coefficients) is the same as the inverse permutation on the coefficients (fixing the $e_i$). If one uses $\pi$ in place of $\pi^{-1}$ in the definition of $\varphi_\pi$ then the map $\pi \mapsto \varphi_\pi$ is not necessarily a homomorphism – it corresponds to a right group action.)

First we show that this mapping $\psi : \pi \mapsto \varphi_\pi$ is a homomorphism. Let $g = (g_i) \in G$.

We have $[(\varphi_{\pi_1} \circ \varphi_{\pi_2})(g)]_i$ $= [\varphi_{\pi_1}(\varphi_{\pi_2}(g))]_i$ $= [\varphi_{\pi_2}(g)]_{\pi_1^{-1}(i)}$ $= g_{\pi_2^{-1}(\pi_1^{-1}(i))}$ $= g_{(\pi_2^{-1} \circ \pi_1^{-1})(i)}$ $= g_{(\pi_1 \circ \pi_2)^{-1}(i)}$ $= [\varphi_{\pi_1 \circ \pi_2}(g)]_i$. Thus $\varphi_{\pi_1} \circ \varphi_{\pi_2} = \varphi_{\pi_1\pi_2}$, hence this mapping is a homomorphism.

Now suppose the $G_i$ are nontrivial and let $\pi$ be in the kernel of $\psi$. Then for all $g \in G$, $g_i = (\varphi_\pi(g))_i = g_{\pi^{-1}}(i)$. Since the $G_i$ are nontrivial, we may choose $g$ such that any given pair of entries consists of distinct elements of $G_i$. Thus $\pi(i) = i$ for all $i$, and in fact $\pi = 1$. Thus the kernel of $\psi$ is trivial, hence $\psi$ is injective.

Note that the nontrivial condition on $G_i$ is necessary because otherwise, $G \cong 1$ and $\mathsf{Aut}(G) \cong 1$.

### An action of a quotient group on an abelian normal subgroup

Let $G$ be a group and let $A$ be an abelian, normal subgroup of $G$. Show that $G/A$ acts (on the left) by conjugation on $A$ by $[g] \cdot a = gag^{-1}$. (In particular, show that this action is well defined.) Give an explicit example to show that this action is not well defined if $A$ is nonabelian.

To see well definedness, suppose $gh^{-1} = b \in A$. That is, $g = bh$. Now $gag^{-1} = bhah^{-1}b^{-1}$; note that $hah^{-1} \in A$ since $A$ is normal, hence $gag^{-1} = hah^{-1}$ since $A$ is abelian. Thus $[g] \cdot a = [h] \cdot a$, and the action is well defined.

Clearly we have $[1] \cdot a = a$ and $[gh] \cdot a = ghah^{-1}g^{-1} = [g] \cdot ([h] \cdot a) = ([g][h]) \cdot a$. So $G/A$ acts on $A$.

Now identify $D_6 = \langle r^2, s \rangle$ in $D_{12}$. Note that $D_6 \leq D_{12}$ has index 2, hence is normal, and that $D_6$ is nonabelian. Moreover, $r$ and $rs$ are distinct representatives of $rD_6$. However, $rr^2r^{-1} = r^2$ and $rsr^2rs = r^4$, so that the action described above is not well defined.

### The orbits of corresponding left and right group actions are the same

Suppose $G$ acts on $A$ on the left via $g \cdot a$. Denote the corresponding right action of $G$ on $A$ by $a \cdot g$; that is, $a \cdot g = g^{-1} \cdot a$. Prove that the induced equivalence relations $\sim_\ell$ and $\sim_r$, defined by $a \sim_\ell b$ if and only if $a = g \cdot b$ for some $g \in G$ and $a \sim_r b$ if and only if $a = b \cdot g$ for some $g \in G$, are the same relation. (That is, prove that $a \sim_\ell b$ if and only if $a \sim_r b$.)

If $a \sim_\ell b$, then $a = g \cdot b$ for some $g \in G$. Then $g^{-1} \cdot a = b$, so that $a \cdot g = b$. Thus $b \sim_r a$, and since $\sim_r$ is reflexive, $a \sim_r b$.

If $a \sim_r b$, then $a = b \cdot g$ for some $g \in G$. Then $a \cdot g^{-1} = b$, so that $g \cdot a = b$. Thus $b \sim_\ell a$, and since $\sim_\ell$ is reflexive, $a \sim_\ell b$.

### Transitive group actions induce transitive actions on the orbits of the action of a subgroup

Suppose $G \leq S_A$ acts transitively on $A$ and let $H \leq G$ be normal. Let $O_1, O_2, \ldots, O_r$ be the distinct orbits of $H$ on $A$.

1. Prove that $G$ permutes the sets $O_k$ in the sense that for each $g \in G$ and each $i \in \{1,2,\ldots,r\}$ there exists a $j$ such that $g \cdot O_i = O_j$. Prove that $G$ is transitive on $\{ O_1,O_2,\ldots,O_r \}$. Deduce that all orbits of $H$ on $A$ have the same cardinality.
2. Prove that if $a \in O_1$ then $|O_1| = [H : H \cap \mathsf{stab}_G(a)]$ and that $r = [G : H \mathsf{stab}_G(a)]$. [Hint: Note that $H \cap \mathsf{stab}_G(a) = \mathsf{stab}_H(a)$ and use the Second Isomorphism Theorem.]

1. Let $g \in G$ and $a \in A$. Now $H \cdot a$ is an arbitrary orbit of $H$ on $A$. Moreover, $g \cdot a = b$ for some $b \in A$. Then $g \cdot (H \cdot a) = gH \cdot a = Hg \cdot a = H \cdot (g \cdot a) = H \cdot b$, since $H$ is normal. Thus $G$ permutes the orbits $\{ H \cdot a \ |\ a \in A \}$.

Now let $a,b \in A$ be arbitrary; since the action of $G$ on $A$ is transitive, there exists $g \in G$ such that $g \cdot a = b$. Then $g \cdot (H \cdot a) = gH \cdot a = Hg \cdot a = H \cdot (g \cdot a) = H \cdot b$; thus the action of $G$ on the orbits $\{ H \cdot a \ |\ a \in A \}$ is transitive.

Let $a,b \in A$ be arbitrary, and let $g \in G$ such that $g \cdot a = b$. Because $H$ is normal, for every $h \in H$ there exists a unique $k \in H$ such that $gh = kg$. We have a mapping $\varphi_g : H \cdot a \rightarrow H \cdot b$ given by $\varphi_g(h \cdot a) = k \cdot b$. Because $gH = Hg$, this mapping $\varphi_g$ is bijective, and $|H \cdot a| = |H \cdot b|$. Because the action of $G$ is transitive on the orbits, all orbits have the same cardinality.

2. Let $a \in A$. We can restrict the action of $G$ on the orbits of $H$ to $H$; with this action we have $|H \cdot a| = [H : \mathsf{stab}_H(a)]$. Clearly $\mathsf{stab}_H(a) = H \cap \mathsf{stab}_G(a)$; thus $|H \cdot a| = [H : H \cap \mathsf{stab}_G(a)]$. Considering now the (transitive) action of $G$ on the orbits of $H$, we have $r = |\{ H \cdot a \ |\ a \in A \}| = [G : \mathsf{stab}_G(H \cdot a)]$.

We claim that $\mathsf{stab}_G(H \cdot a) = H \mathsf{stab}_G(a)$. Proof of claim: Suppose $g \in \mathsf{stab}_G(H \cdot a)$. Now $Hg \cdot a = gH \cdot a = g \cdot (H \cdot a) = H \cdot a$, so that in particular $g \cdot a = h \cdot a$ for some $h \in H$. Then $h^{-1}g \cdot a = a$, so that $h^{-1}g \in \mathsf{stab}_G(a)$. Thus $g \in H \mathsf{stab}_G(a)$. Now suppose $g \in H \mathsf{stab}_G(a)$. Then $g = hx$ for some $h \in H$ and $x \in \mathsf{stab}_G(a)$. Now $g \cdot (H \cdot a) = gH \cdot a = Hg \cdot a = Hhx \cdot a = H \cdot a$, so that $g \in \mathsf{stab}_G(H \cdot a)$. Thus $r = [G : H \mathsf{stab}_G(a)]$.

### Basic properties of blocks of a group action

Let $G \leq S_A$ act transitively on the set $A$. A block is a nonempty subset $B \subseteq A$ such that for all $\sigma \in G$ either $\sigma[B] = B$ or $\sigma[B] \cap B = \emptyset$.

1. Prove that if $B$ is a block containing $a \in A$ and we define $\mathsf{stab}(B) = \{ \sigma \in G \ |\ \sigma[B] = B \}$ then $\mathsf{stab}(a) \leq \mathsf{stab}(B) \leq G$.
2. Prove that if $B$ is a block then $\mathcal{B} = \{ \sigma[B] \ |\ \sigma \in G \}$ is a partition of $A$.
3. A (transitive) group $G \leq S_A$ is called primitive if the only blocks in $A$ are the trivial ones- the singletons and $A$ itself. Prove that $S_4$ is primitive on $A = \{1,2,3,4\}$. Prove that $D_8$ is not primitive as a permutation group on the four vertices of a square.
4. Let $G \leq S_A$ act transitively on $A$. Prove that $G$ is primitive on $A$ if and only if for all $a \in A$, $\mathsf{stab}(a)$ is maximal in $G$.

1. Note that $\mathsf{stab}(B)$ is not empty since $1[B] = B$. Now let $\sigma, \tau \in \mathsf{stab}(B)$. Note that $\tau^{-1}[B] = \tau^{-1}\tau [B] = B$, so that $\sigma\tau^{-1}[B] = B$. Thus $\sigma\tau^{-1} \in \mathsf{stab}(B)$. By the subgroup criterion, $\mathsf{stab}(B) \leq G$.

Now suppose $\sigma \in \mathsf{stab}(a)$. We have $\sigma(a) = a$, so that $\{a\} \subseteq \sigma[B] \cap B$. Hence $\sigma[B] \cap B$ is nonempty, and we have $\sigma[B] = B$ since $B$ is a block. Thus $\sigma \in \mathsf{stab}(B)$. By a previous exercise, $\mathsf{stab}(a) \leq \mathsf{stab}(B)$.

2. First we show that $\bigcup_{\sigma \in G} \sigma[B] = A$. The $(\subseteq)$ direction is clear. $(\supseteq)$: Let $a \in A$ and $b \in B$. Since the action of $G$ is transitive, there exists $\sigma \in G$ such that $a = \sigma(b)$. Then $a \in \sigma[B]$, so that $aA \subseteq \bigcup_{\sigma \in G} \sigma[B]$.

Now suppose $\sigma[B] \cap \tau[B] \neq \emptyset$. Then there exist $a,b \in B$ such that $\sigma(a) = \tau(b)$. Now $a = \sigma^{-1}\tau \cdot b$, so that $\sigma^{-1}\tau[B] \cap B$ is not empty. Thus $\sigma^{-1}\tau[B] = B$, and we have $\sigma[B] = \tau[B]$. So elements of $\mathcal{B}$ are pairwise disjoint; hence $\mathcal{B}$ is a partition of $A$.

1. Let $B \subseteq A = \{1,2,3,4\}$ be a proper nonempty subset. Then there exist elements $x \in B$ and $y \in A \setminus B$. Suppose $B$ is a block. Consider $\sigma = (x\ y) \in S_4$; since $\sigma(x) \notin B$, we have $\sigma[B] \cap B = \emptyset$. Let $w$ and $z$ be the remaining elements of $A$. If $w \in B$, then $w \in \sigma[B] \cap B$, a contradiction; similarly for $z$. Thus $B = \{x\}$. Now clearly $A$ itself is a block, and any proper block must be a singleton. Thus this action of $S_4$ is primitive.
2. We saw previously that $D_8 \cong \langle (1\ 3), (1\ 2\ 3\ 4) \rangle$, where $A = \{1,2,3,4\}$ are labels on the vertices of a square (written clockwise). Consider the set $\{1,3\}$. We see that
1. $1 \cdot \{1,3\} = \{1,3\}$
2. $(1\ 2\ 3\ 4) \cdot \{1,3\} = \{2,4\}$
3. $(1\ 3)(2\ 4) \cdot \{1,3\} = \{1,3\}$
4. $(1\ 4\ 3\ 2) \cdot \{1,3\} = \{2,4\}$
5. $(1\ 3) \cdot \{1,3\} = \{1,3\}$
6. $(1\ 2)(3\ 4) \cdot \{1,3\} = \{2,4\}$
7. $(2\ 4) \cdot \{1,3\} = \{1,3\}$
8. $(1\ 4)(2\ 3) \cdot \{1,3\} = \{2,4\}$

So that $\{1,3\}$ is a nontrivial block; hence this action is not primitive.

3. We begin with some lemmas.

Lemma 1: Let $G \leq S_A$ act transitively on $A$ and let $B \subseteq A$ be a block under the action. Then $\mathsf{stab}(B) = G$ if and only if $B = A$. Proof: The $(\Leftarrow)$ direction is clear. $(\Rightarrow)$ Suppose $B$ is a proper subset and let $x \in B$, $y \in A \setminus B$. Now $y \in (x\ y)[B]$ and $y \notin B$, so that $(x\ y)[B] \neq B$. Since $B$ is a block, we have $(x\ y)[B] \cap B = \emptyset$. But $x \in (x\ y)[B] \cap B$ since $x \in B$ and $y \notin B$, a contradiction. So $B$ is not proper, and we have $A = B$. $\square$

Now we move to the main result.

$(\Rightarrow)$ Suppose $G$ is primitive on $A$; then the only blocks of $A$ are $A$ and the singletons. Now let $a \in A$ and let $H \leq G$ be a subgroup with $\mathsf{stab}(a) \leq H \leq G$. We consider the set $H \cdot a = \{ h \cdot a \ |\ h \in H \}$.

We claim that $H \cdot a$ is a block. Proof of claim: $H \cdot a$ is not empty since $a = 1 \cdot a \in H \cdot a$. Now let $\sigma \in G$. If $\sigma \in H$, then $\sigma \cdot (H \cdot a) = \sigma H \cdot a = H \cdot a$. If $\sigma \notin H$, then $\sigma \cdot (H \cdot a) = \sigma H \cdot a$. Suppose that $\sigma H \cdot a \cap H \cdot a \neq \emptyset$; say that $\sigma\tau_1 \cdot a = \tau_2 \cdot a$ for some $\tau_1, \tau_2 \in H$. Then $\tau_2^{-1} \sigma \tau_1 \cdot a = a$, so that $\tau_2^{-1} \sigma \tau_1 \in \mathsf{stab}(a) \leq H$. But then $\sigma \in \tau_2H\tau_1^{-1} = H$, a contradiction. Thus if $\sigma \notin H$, then $\sigma[H \cdot a] \cap H \cdot a = \emptyset$. Hence $H \cdot a$ is a block.

Next we claim that $\mathsf{stab}(H \cdot a) = H$. Proof of claim: $(\supseteq)$ If $\tau \in H$, then $\tau \cdot (H \cdot a) = \tau H \cdot a = H \cdot a$. $(\subseteq)$ Suppose $\sigma \cdot (H \cdot a) = \sigma H \cdot a = H \cdot a$. Then $\sigma \cdot a = \tau \cdot a$ for some $\tau \in H$, hence $\tau^{-1} \sigma \cdot a = a$. Then $\tau^{-1} \sigma \in \mathsf{stab}(a) \leq H$, so that $\sigma \in H$.

Since $G$ is primitive on $a$ and $a \in H \cdot a$, we have $H \cdot a = \{a\}$ or $H \cdot a = A$. If $H \cdot a = \{a\}$, we have $H \leq \mathsf{stab}(a)$, so that $H = \mathsf{stab}(a)$. If $H \cdot a = A$, we have $H = \mathsf{stab}(H \cdot a) = \mathsf{stab}(A) = G$. Thus $\mathsf{stab}(a)$ is a maximal subgroup of $G$.

$(\Leftarrow)$ Suppose that for all $a \in A$, $\mathsf{stab}(a)$ is a maximal subgroup in $G$. Let $B \subseteq A$ be a block with $a \in B$. Now $\mathsf{stab}(a) \leq \mathsf{stab}(B) \leq G$ by part 1 above. Since $\mathsf{stab}(a)$ is maximal, there are two cases.

If $\mathsf{stab}(B) = \mathsf{stab}(a)$ and $b \in B$ such that $b \neq a$, then since $G$ acts transitively there exists $\sigma \in G$ such that $\sigma \cdot a = b$. Now $\sigma \in \mathsf{stab}(B) = \mathsf{stab}(a)$, a contradiction. Thus $B = \{a\}$.

If $\mathsf{stab}(B) = G$, then by Lemma 1 we have $B = A$.

Thus $G$ is primitive on $A$.

### Compute some orbits of an action by Sym(4) on polynomials in four variables

As in this previous exercise, let $S_4$ act on the set $R$ of all polynomials with integer coefficients in the independent variables $x_1,x_2,x_3,x_4$ by permuting the indices: $\sigma \cdot p(x_1,x_2,x_3,x_4) = p(x_{\sigma(1)}, x_{\sigma(2)}, x_{\sigma(3)}, x_{\sigma(4)})$.

1. Find the polynomials in the orbit of $x_1 + x_2$. (Recall from a previous exercise that the stabilizer of this element has order 4.)
2. Find the polynomials in the orbit of $x_1x_2 + x_3x_4$. (Recall that the stabilizer of this element has order 8.)
3. Find the polynomials in the orbit of $(x_1 + x_2)(x_3 + x_4)$.

1. We have $[S_4 : \mathsf{stab}(x_1+x_2)] = 6$. A simple calculation then shows that this orbit is $\{ x_1+x_2, x_1+x_3, x_1+x_4,$ $x_2+x_3, x_2+x_4, x_3+x_4 \}$.
2. We have $[S_4 : \mathsf{stab}(x_1x_2 + x_3x_4)] = 3$. A simple calculation then shows that this orbit is $\{ x_1x_2+x_3x_4, x_1x_3+x_2x_4, x_1x_4+x_2x_3 \}$.
3. We saw in a previous exercise that $\mathsf{stab}((x_1+x_2)(x_3+x_4)) = \mathsf{stab}(x_1x_2+x_3x_4)$. Thus $[S_4 : \mathsf{stab}((x_1+x_2)(x_3+x_4))] = 3$. A simple calculation then shows that this orbit is $\{ (x_1+x_2)(x_3+x_4),$ $(x_1+x_3)(x_2+x_4),$ $(x_1+x_4)(x_2+x_3) \}$.

### Compute the orbits, cycle decompositions, and stabilizers of some group actions of Sym(3)

Repeat the previous example with each of the following actions.

1. $S_3$ acting on the set $A = \{ (i,j,k) \ |\ 1 \leq i,j,k \leq 3 \}$ by $\sigma \cdot (i,j,k) = (\sigma(i), \sigma(j), \sigma(k))$
2. $S_3$ acting on the set $B$ of all nonempty subsets of $\{1,2,3\}$ by $\sigma \cdot X = \sigma[X]$.

• For $S_3$ acting on $A$:
• We compute the orbits as follows.
• $1 \cdot (1,1,1) = (2\ 3) \cdot (1,1,1) = (1,1,1)$, $(1\ 2) \cdot (1,1,1) = (1\ 2\ 3) \cdot (1,1,1) = (2,2,2)$, and $(1\ 3) \cdot (1,1,1) = (1\ 3\ 2) \cdot (1,1,1) = (3,3,3)$, so that $\mathcal{O}_1 = \{ (1,1,1), (2,2,2), (3,3,3) \}$.
• $1 \cdot (1,1,2) = (1,1,2)$, $(1\ 2) \cdot (1,1,2) = (2,2,1)$, $(1\ 3) \cdot (1,1,2) = (3,3,2)$, $(2\ 3) \cdot (1,1,2) = (1,1,3)$, $(1\ 2\ 3) \cdot (1,1,2) = (2,2,3)$ and $(1\ 3\ 2) \cdot (1,1,2) = (3,3,1)$, so that $\mathcal{O}_2 = \{ (1,1,2), (2,2,1), (3,3,2),$ $(1,1,3), (2,2,3), (3,3,1) \}$.
• $1 \cdot (1,2,1) = (1,2,1)$, $(1\ 2) \cdot (1,2,1) = (2,1,2)$, $(2\ 3) \cdot (1,2,1) = (1,3,1)$, $(1\ 2\ 3) \cdot (1,2,1) = (2,3,2)$, and $(1\ 3\ 2) \cdot (1,2,1) = (3,1,3)$, so that $\mathcal{O}_3 = \{ (1,2,1), (2,1,2), (3,2,3),$ $(1,3,1), (2,3,2), (3,1,3) \}$.
• $1 \cdot (2,1,1) = (2,1,1)$, $(1\ 2) \cdot (2,1,1) = (1,2,2)$, $(1\ 3) \cdot (2,1,1) = (2,3,3)$, $(2\ 3) \cdot (2,1,1) = (3,1,1)$, $(1\ 2\ 3) \cdot (2,1,1) = (3,2,2)$, and $(1\ 3\ 2) \cdot (2,1,1) = (1,3,3)$, so that $\mathcal{O}_4 = \{ (2,1,1), (1,2,2),$ $(2,3,3), (3,1,1),$ $(3,2,2), (1,3,3) \}$.
• $1 \cdot (1,2,3) = (1,2,3)$, $(1\ 2) \cdot (1,2,3) = (2,1,3)$, $(1\ 3) \cdot (1,2,3) = (3,2,1)$, $(2\ 3) \cdot (1,2,3) = (1,3,2)$, $(1\ 2\ 3) \cdot (1,2,3) = (2,3,1)$, and $(1\ 3\ 2) \cdot (1,2,3) = (3,1,2)$, so that $\mathcal{O}_5 = \{ (1,2,3), (2,1,3),$ $(3,2,1), (1,3,2),$ $(2,3,1), (3,1,2) \}$.

This exhausts the elements of $A$.

• We fix the labeling
 $\alpha_{1} = (1,1,1)$ $\alpha_{10} = (2,1,1)$ $\alpha_{19} = (3,1,1)$ $\alpha_{2} = (1,1,2)$ $\alpha_{11} = (2,1,2)$ $\alpha_{20} = (3,1,2)$ $\alpha_{3} = (1,1,3)$ $\alpha_{12} = (2,1,3)$ $\alpha_{21} = (3,1,3)$ $\alpha_{4} = (1,2,1)$ $\alpha_{13} = (2,2,1)$ $\alpha_{22} = (3,2,1)$ $\alpha_{5} = (1,2,2)$ $\alpha_{14} = (2,2,2)$ $\alpha_{23} = (3,2,2)$ $\alpha_{6} = (1,2,3)$ $\alpha_{15} = (2,2,3)$ $\alpha_{24} = (3,2,3)$ $\alpha_{7} = (1,3,1)$ $\alpha_{16} = (2,3,1)$ $\alpha_{25} = (3,3,1)$ $\alpha_{8} = (1,3,2)$ $\alpha_{17} = (2,3,2)$ $\alpha_{26} = (3,3,2)$ $\alpha_{9} = (1,3,3)$ $\alpha_{18} = (2,3,3)$ $\alpha_{27} = (3,3,3)$.

Now under the permutation representation $S_3 \rightarrow S_{27}$, we have

• $1 \mapsto 1$
• $(1\ 2) \mapsto (\alpha_1\ \alpha_{14})$ $(\alpha_2\ \alpha_{13})$ $(\alpha_3\ \alpha_{15})$ $(\alpha_4\ \alpha_{11})$ $(\alpha_5\ \alpha_{10})$ $(\alpha_6\ \alpha_{12})$ $(\alpha_7\ \alpha_{17})$ $(\alpha_8\ \alpha_{16})$ $(\alpha_9\ \alpha_{18})$ $(\alpha_{19}\ \alpha_{23})$ $(\alpha_{20}\ \alpha_{22})$ $(\alpha_{21}\ \alpha_{24})$ $(\alpha_{25}\ \alpha_{26})$
• $(1\ 3) \mapsto (\alpha_1\ \alpha_{27})$ $(\alpha_2\ \alpha_{26})$ $(\alpha_3\ \alpha_{25})$ $(\alpha_4\ \alpha_{24})$ $(\alpha_5\ \alpha_{23})$ $(\alpha_6\ \alpha_{22})$ $(\alpha_7\ \alpha_{21})$ $(\alpha_8\ \alpha_{20})$ $(\alpha_9\ \alpha_{19})$ $(\alpha_{10}\ \alpha_{18})$ $(\alpha_{11}\ \alpha_{17})$ $(\alpha_{12}\ \alpha_{16})$ $(\alpha_{13}\ \alpha_{15})$
• $(2\ 3) \mapsto (\alpha_2\ \alpha_3)$ $(\alpha_4\ \alpha_7)$ $(\alpha_5\ \alpha_{9})$ $(\alpha_6\ \alpha_{8})$ $(\alpha_{10}\ \alpha_{19})$ $(\alpha_{11}\ \alpha_{21})$ $(\alpha_{12}\ \alpha_{20})$ $(\alpha_{13}\ \alpha_{25})$ $(\alpha_{14}\ \alpha_{27})$ $\alpha_{15}\ \alpha_{26})$ $(\alpha_{16}\ \alpha_{22})$ $(\alpha_{17}\ \alpha_{24})$ $(\alpha_{18}\ \alpha_{23})$
• $(1\ 2\ 3) \mapsto (\alpha_1\ \alpha_{14}\ \alpha_{27})$ $(\alpha_2\ \alpha_{15}\ \alpha_{25})$ $(\alpha_3\ \alpha_{13}\ \alpha_{26})$ $(\alpha_{4}\ \alpha_{17}\ \alpha_{21})$ $(\alpha_5\ \alpha_{18}\ \alpha_{19})$ $(\alpha_6\ \alpha_{16}\ \alpha_{20})$ $(\alpha_7\ \alpha_{11}\ \alpha_{24})$ $(\alpha_8\ \alpha_{12}\ \alpha_{22})$ $(\alpha_9\ \alpha_{10}\ \alpha_{23})$
• $(1\ 3\ 2) \mapsto (\alpha_1\ \alpha_{27}\ \alpha_{14})$ $(\alpha_2\ \alpha_{25}\ \alpha_{15})$ $(\alpha_3\ \alpha_{26}\ \alpha_{13})$ $(\alpha_{4}\ \alpha_{21}\ \alpha_{17})$ $(\alpha_5\ \alpha_{19}\ \alpha_{18})$ $(\alpha_6\ \alpha_{20}\ \alpha_{16})$ $(\alpha_7\ \alpha_{24}\ \alpha_{11})$ $(\alpha_8\ \alpha_{22}\ \alpha_{12})$ $(\alpha_9\ \alpha_{23}\ \alpha_{10})$
• If $x \in \mathcal{O}_1$, then $[S_3 : \mathsf{stab}(x)] = 3$ so that $|\mathsf{stab}(x)| = 2$. In particular, consider $(3,3,3) \in \mathcal{O}_1$. Since $(1\ 2) \cdot (3,3,3) = (3,3,3)$, we have $\mathsf{stab}((3,3,3)) = \langle (1\ 2) \rangle$.

Each remaining orbit has order 6. Thus, if $x \in \mathcal{O}_k$, $2 \leq k \leq 5$, then $|\mathsf{stab}(x)| = 1$; hence $\mathsf{stab}(x) = 1$.

• For $S_3$ acting on $B$:
• We compute the orbits as follows.
• We have $1 \cdot \{1\} = (2\ 3) \cdot \{1\} = \{1\}$, $(1\ 2) \cdot \{1\} = (1\ 2\ 3) \cdot \{1\} = \{2\}$, and $(1\ 3) \cdot \{1\} = (1\ 3\ 2) \cdot \{1\} = \{3\}$. Thus $\mathcal{O}_1 = \{ \{1\}, \{2\}, \{3\} \}$.
• We have $1 \cdot \{1,2\} = (1\ 2) \cdot \{1,2\} = \{1,2\}$, $(1\ 3) \cdot \{1,2\} = (1\ 2\ 3) \cdot \{1,2\} = \{2,3\}$, and $(2\ 3) \cdot \{1,2\} = (1\ 3\ 2) \cdot \{1,2\} = \{1,3\}$. Thus $\mathcal{O}_2 = \{ \{1,2\}, \{1,3\}, \{2,3\} \}$.
• There is only one remaining element of $B$. Thus $\mathcal{O}_3 = \{ \{1,2,3\} \}$.
• We fix the labeling
 $\alpha_{1} = \{1\}$ $\alpha_{5} = \{1,3\}$ $\alpha_{2} = \{2\}$ $\alpha_{6} = \{2,3\}$ $\alpha_{3} = \{3\}$ $\alpha_{7} = \{1,2,3\}$ $\alpha_{4} = \{1,2\}$

Under the permutation representation $S_3 \rightarrow S_7$,

• $1 \mapsto 1$
• $(1\ 2) \mapsto (\alpha_1\ \alpha_2)(\alpha_5\ \alpha_6)$
• $(1\ 3) \mapsto (\alpha_1\ \alpha_3)(\alpha_4\ \alpha_6)$
• $(2\ 3) \mapsto (\alpha_2\ \alpha_3)(\alpha_4\ \alpha_5)$
• $(1\ 2\ 3) \mapsto (\alpha_1\ \alpha_2\ \alpha_3)(\alpha_4\ \alpha_6\ \alpha_5)$
• $(1\ 3\ 2) \mapsto (\alpha_1\ \alpha_3\ \alpha_2)(\alpha_4\ \alpha_5\ \alpha_6)$
• If $x \in \mathcal{O}_1$, then $[S_3 : \mathsf{stab}(x)] = 3$, so that $|\mathsf{stab}(x)| = 2$. In particular, since $(2\ 3) \cdot \{1\} = \{1\}$, we have $\mathsf{stab}(\{1\}) = \langle (2\ 3) \rangle$.

If $x \in \mathcal{O}_2$, we similarly have $|\mathsf{stab}(x)| = 2$. In particular, since $(2\ 3) \cdot \{2,3\} = \{2,3\}$, we have $\mathsf{stab}(\{2,3\}) = \langle (2\ 3) \rangle$.

Now $[S_3 : \mathsf{stab}(\{1,2,3\})] = 1$, so that $\mathsf{stab}(\{1,2,3\}) = S_3$.

### Compute the orbits, cycle decompositions, and stabilizers of some given group actions of Sym(3)

Let $S_3$ act on the set $A = \{ (i,j) \ |\ 1 \leq i,j \leq 3 \}$ by $\sigma \cdot (i,j) = (\sigma(i), \sigma(j))$.

1. Find the orbits of $S_3$ on $A$.
2. Fix a labeling of the elements of $A$. For each $\sigma \in S_3$, find the cycle decomposition of $\sigma$ under the permutation representation $S_3 \rightarrow S_9$.
3. For each orbit $\mathcal{O}$ of this action, choose some $a \in \mathcal{O}$ and compute $\mathsf{stab}(a)$.

1. We have $1 \cdot (1,1) = (2\ 3) \cdot (1,1) = (1,1)$, $(1\ 2) \cdot (1,1) = (1\ 2\ 3) \cdot (1,1) = (2,2)$, and $(1\ 3) \cdot (1,1) = (1\ 3\ 2) \cdot (1,1) = (3,3)$. Thus the orbit containing $(1,1)$ is $\mathcal{O}_1 = \{ (1,1), (2,2), (3,3) \}$.

We also have $1 \cdot (1,2) = (1,2)$, $(1\ 2) \cdot (1,2) = (2,1)$, $(1\ 3) \cdot (1,2) = 3,2)$, $(2\ 3) \cdot (1,2) = (1,3)$, $(1\ 2\ 3) \cdot (1,2) = (2,3)$, and $(1\ 3\ 2) \cdot (1,2) = (3,2)$. Thus the orbit containing $(1,2)$ is $\mathcal{O}_2 = \{ (1,2), (2,1), (3,2),$ $(1,3), (2,3), (3,2) \}$. These two orbits exhaust the elements of $A$.

2. Fix the labeling
 $\alpha_{1} = (1,1)$ $\alpha_{4} = (2,1)$ $\alpha_{7} = (3,1)$ $\alpha_{2} = (1,2)$ $\alpha_{5} = (2,2)$ $\alpha_{8} = (3,2)$ $\alpha_{3} = (1,3)$ $\alpha_{6} = (2,3)$ $\alpha_{9} = (3,3)$

Under the permutation representation $S_3 \rightarrow S_9$, we have

1. $1 \mapsto 1$
2. $(1\ 2) \mapsto (\alpha_1\ \alpha_5)(\alpha_2\ \alpha_4)(\alpha_3\ \alpha_6)(\alpha_7\ \alpha_8)$
3. $(1\ 3) \mapsto (\alpha_1\ \alpha_9)(\alpha_2\ \alpha_8)(\alpha_3\ \alpha_7)(\alpha_4\ \alpha_6)$
4. $(2\ 3) \mapsto (\alpha_2\ \alpha_3)(\alpha_4\ \alpha_7)(\alpha_5\ \alpha_9)(\alpha_6\ \alpha_8)$
5. $(1\ 2\ 3) \mapsto (\alpha_1\ \alpha_5\ \alpha_9)(\alpha_2\ \alpha_6\ \alpha_7)(\alpha_3\ \alpha_4\ \alpha_8)$
6. $(1\ 3\ 2) \mapsto (\alpha_1\ \alpha_9\ \alpha_5)(\alpha_2\ \alpha_7\ \alpha_6)(\alpha_3\ \alpha_8\ \alpha_4)$
3. If $\sigma \in \mathcal{O}_1$, we have $[S_3 : \mathsf{stab}(\sigma)] = 3$, so that $|\mathsf{stab}(\sigma)| = 2$. Consider $(2,2) \in \mathcal{O}_1$; since $(1\ 3) \cdot (2,2) = (2,2)$, we have $\mathsf{stab}((2,2)) = \langle (1\ 3) \rangle$.

If $\sigma \in \mathcal{O}_2$, then $[S_3 : \mathsf{stab}(\sigma)] = 6$. Hence $|\mathsf{stab}(\sigma)| = 1$, and we have $\mathsf{stab}(\sigma) = 1$.