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.