Tag Archives: symmetric group

Compute in a group ring

Consider the following elements of the group ring \mathbb{Z}/(3)[S_3]: \alpha = 1(2\ 3) + 2(1\ 2\ 3) and \beta = 2(2\ 3) + 2(1\ 3\ 2). Compute \alpha + \beta, 2\alpha - 3\beta, \alpha\beta, \beta\alpha, and \alpha^2.


Evidently,

  1. \alpha + \beta = 2(1\ 2\ 3) + 2(1\ 3\ 2)
  2. 2\alpha - 3\beta = 2\alpha = 2(2\ 3) + (1\ 2\ 3)
  3. \alpha\beta = 2(1\ 2) + 1(1\ 2\ 3)
  4. \beta\alpha = 1(1\ 3) + 2(1\ 3\ 2)
  5. \alpha^2 = 1(1) + 2(1\ 2) + 2(1\ 3) + 1(1\ 3\ 2)

Compute in a group ring

Consider the following elements of the integral group ring \mathbb{Z}[S_3]: \alpha = 3(1\ 2) - 5(2\ 3) + 14(1\ 2\ 3) and \beta = 6(1) + 2(2\ 3) - 7(1\ 3\ 2). Compute the following elements: \alpha + \beta, 2\alpha - 3\beta, \alpha\beta, \beta\alpha, and \alpha^2.


Evidently,

  1. \alpha + \beta = 6(1) + 3(1\ 2) - 3(2\ 3) + 14(1\ 2\ 3) - 7(1\ 3\ 2)
  2. 2\alpha - 3\beta = -18(1) + 6(1\ 2) - 16(2\ 3) + 28(1\ 2\ 3) + 21(1\ 3\ 2)
  3. \alpha\beta = -108(1) + 81(1\ 2) - 21(1\ 3) - 30(2\ 3) + 90(1\ 2\ 3)
  4. \beta\alpha = -108(1) + 18(1\ 2) + 63(1\ 3) - 51(2\ 3) + 84(1\ 2\ 3) + 6(1\ 3\ 2)
  5. \alpha^2 = 34(1) - 70(1\ 2) - 28(1\ 3) + 42(2\ 3) - 15(1\ 2\ 3) + 181(1\ 3\ 2)

An automorphism of Sym(6) that is not inner

This exercise exhibits an automorphism of S_6 that is not inner (hence it shows that [\mathsf{Aut}(S_6) : \mathsf{Inn}(S_6)] = 2. Let u_1 = (1\ 2)(3\ 4)(5\ 6), u_2 = (1\ 4)(2\ 5)(3\ 6), u_3 = (1\ 3)(2\ 4)(5\ 6), u_4 = (1\ 2)(3\ 6)(4\ 5), and u_5 = (1\ 4)(2\ 3)(5\ 6). Show that u_1, \ldots, u_5 satisfy the following relations:

  • u_i^2 = 1 for all i,
  • (u_iu_j)^2 = 1 for all i,j with |i-j| \geq 2, and
  • (u_iu_{i+1})^3 = 1 for all i \in \{1,2,3,4\}.

Deduce that S_6 = \langle u_1, \ldots, u_5 \rangle and that the map (1\ 2) \mapsto u_1, (2\ 3) \mapsto u_2, (3\ 4) \mapsto u_3, (4\ 5) \mapsto u_4, (5\ 6) \mapsto u_5 extends to an automorphism of S_6 (which is clearly not inner since it does not preserve cycle shape.


Showing that u_i satisfy the given relations is a straightforward computation. Moreover, they generate S_6 since u_3u_5u_1 = (6\ 5) and u_1u_2u_4 = (6\ 5\ 2\ 3\ 4\ 1).

The automorphism group of the quaterion group is isomorphic to Sym(4)

Prove that \mathsf{Aut}(Q_8) \cong S_4.


We already know that |\mathsf{Aut}(Q_8)| = 24 by counting the possible images of, say, i and j. By a previous theorem, it suffices to show that no automorphism of Q_8 has order 6.

Let \varphi be an automorphism of Q_8. Now Q_8 has 6 elements of order 4, one element of order 2, and one identity; \varphi must fix the identity and the element of order 2, so that we can consider \varphi as an element of S_A, where A = \{i,j,k,-i,-j,-k\}. Suppose now that some automorphism \varphi has order 6. Now \varphi(i) = a, where a \notin \{i,-i\}, and \varphi(a) = b, where b \notin \{i,-i,a,-a\}. Clearly \varphi(-i) = -a and \varphi(-a) = -b. If \varphi(b) = i, then \varphi = (i\ a\ b)(-i\ -a\ -b) and \varphi has order 3; thus \varphi(b) = -i, and we have \varphi = (i\ a\ b\ -i\ -a\ -b). Now \varphi^3 = (i\ -i)(a\ -a)(b\ -b). Note that ia \in \{b,-b\}. If ia = b, then ia = (-i)(-a) = \varphi^3(i)\varphi^3(a) = \varphi^3(ia) = \varphi^3(b) = -b. Similarly, if ia = -b, then ia = b. Both cases yield a contradiction, so that no automorphism of order 6 exists.

In addition, we record this handy lemma.

Lemma: Let G be a group and let K \leq G be characteristic. Then \mathsf{Aut}(G) acts on G/K by \varphi \cdot xK = \varphi(x)K. Proof: First we need to show well-definedness: if xK = yK, we have xy^{-1} \in K. Then \varphi(xy^{-1}) = \varphi(x)\varphi(y)^{-1} \in K since K is characteristic in G, so that \varphi(x)K = \varphi(y)K. Moreover, 1 \cdot xK = 1(x)K = xK and (\varphi \circ \psi) \cdot xK = (\varphi \circ \psi)(x)K = \varphi(\psi(x))K = \varphi \cdot (\psi(x)K) = \varphi \cdot (\psi \cdot xK). \square

Exhibit a presentation for Sym(4) with two generators

Establish a finite presentation for S_4 with 2 generators.


Recall that S_4 is generated by \alpha = (1\ 2) and \beta = (1\ 2\ 3\ 4). Evidently, these elements satisfy the relations \alpha^2 = \beta^4 = (\alpha\beta)^3 = 1. We will now show that any group presented by \langle a, b \ |\ a^2 = b^4 = (ab)^3 \rangle has at most 24 elements.

  • There is one reduced word of length 0: 1.
  • There are two reduced words of length 1: a and b.
  • There are at most four reduced words of length 2: a^2 = 1, ab, ba, and b^2. We see that at most three of these are reduced.
  • There are at most six reduced words of length 3: aba, ab^2, ba^2 = b, bab, b^2a, b^3. We see that at most five of these are reduced.
  • There are at most ten reduced words of length 4: aba^2 = ab, abab, ab^2a, ab^3, baba = ab^3, bab^2, b^2a^2 = b^2, b^2ab, b^3a = abab, and b^4 = 1. At most five of these are reduced.
  • There are at most ten reduced words of length 5: ababa = b^3, abab^2, ab^2a^2 = ab^2, ab^2ab, ab^3a = bab, ab^4 = a, bab^2a, bab^3, b^2aba = bab^3, and b^2ab^2. We see that at most five of these are reduced.
  • There are at most ten reduced words of length 6: abab^2a, abab^3, ab^2aba = abab^3, ab^2ab^2, bab^2a^2 = bab^2, bab^2ab = ab^2a, bab^3a = b^2ab, bab^4 = ba, b^2ab^2a = ab^2ab^2, and b^2ab^3 = abab^2a. We see that at most three of these are reduced.
  • There are at most six reduced words of length 7: abab^2a^2 = abab^2, abab^2ab = b^2a, abab^3a = ab^2ab, abab^4 = aba, ab^2ab^2a = b^2ab^2, and ab^2ab^3 = ab^2ab. We see that none of these is reduced.

Thus any group with this presentation has at most 24 elements, and in fact we have S_4 = \langle a,b \ |\ a^2 = b^4 = (ab)^3 = 1 \rangle.

A group of order 24 with no elements of order 6 is isomorphic to Sym(4)

Show that a group of order 24 with no element of order 6 is isomorphic to S_4.


Note that 24 = 2^3 \cdot 3, so that Sylow’s Theorem forces n_2(G) \in \{1,3\} and n_3(G) \in \{1,4\}.

Suppose n_3(G) = 1. If n_2(G) = 1, then by the recognition theorem for direct products, G \cong P_2 \times P_3, where P_2 and P_3 are Sylow 2- and 3-subgroups of G, respectively. By Cauchy, there exist elements x \in P_2 and y \in P_3 of order 2 and 3, so that xy has order 6, a contradiction. Suppose now that n_2(G) = 3. Since n_2(G) \not\equiv 1 mod 4, there exist P_2,Q_2 \in \mathsf{Syl}_2(G) such that P_2 \cap Q_2 is nontrivial and its normalizer has order 2^3 \cdot 3, by a previous theorem. Thus P_2 \cap Q_2 \leq G is normal. Note that |P_2 \cap Q_2| is either 2 or 4.

Suppose |P_2 \cap Q_2| = 4. Then at most 7 + 3 + 7 nonidentity elements of G are contained in Sylow 2-subgroups, and 3 elements are contained in Sylow 3-subgroups. This leaves 4 elements not of prime power order, one of which must have order 6- a contradiction.

Suppose now that |P_2 \cap Q_2| = 2. By the N/C Theorem, G/C_G(P_2 \cap Q_2) \leq \mathsf{Aut}(P_2 \cap Q_2) \cong 1, so that P_2 \cap Q_2 is central in G. By Cauchy, there exist elements x \in P_2 \cap Q_2 and y \in G of order 2 and 3, so that xy has order 6, a contradiction.

Thus we may assume n_3(G) = 4. Let P_3 \leq G be a Sylow 3-subgroup and let N = N_G(P_3). The action of G on G/N yields a permutation representation G \rightarrow S_4 whose kernel K is contained in N. Recall that normalizers of Sylow subgroups are self normalizing, so that N is not normal in G. Moreover, we have |N| = 6. We know from the classification of groups of order 6 that N is isomorphic to either Z_6 or D_6; however, in the first case we have an element of order 6, a contradiction. Thus N \cong D_6. We know also that the normal subgroups of D_6 have order 1, 3, or 6. If |K| = 6, then K = N is normal in G, a contradiction. If |K| = 3, then by the N/C theorem we have G/C_G(K) \leq \mathsf{Aut}(Z_3) \cong Z_2. In particular, C_G(K) contains an element of order 2, so that G contains an element of order 6, a contradiction.

Thus K = 1, and in fact G \leq S_4. Since |G| = |S_4| = 24 is finite, G \cong S_4.

Computation of some Frattini subgroups

Compute \Phi(S_3), \Phi(A_4), \Phi(S_4), \Phi(A_5), and \Phi(S_5).


We begin with some lemmas.

Lemma 1: If H \leq S_4 is a subgroup of order 12, then H = A_4. Proof: By Cauchy, H contains an element of order 2 and an element of order 3. If some element of order 2 is a product of two 2-cycles, then A_4 \leq H, and by counting elements we have H = A_4. Suppose no element in H is a product of two 2-cycles; then also no element of H is a 4-cycle, since the square of a 4-cycle is a product of two 2-cycles. Suppose now that H contains two 3-cycles which are not inverses of one another; from the subgroup lattice of A_4, then, A_4 \leq H; since H also contains a 2-cycle, we have H = S_4, a contradiction. Suppose then that H contains exactly two 3-cycles, (a\ b\ c) and (a\ c\ b). then H must contain nine 2-cycles, a contradiction since S_4 contains only six 2-cycles. Thus we see that H = A_4. \square

Lemma 2: If H \leq S_4 has order 6, then H is maximal. Proof: If H is not maximal, then H is contained in some subgroup K of order 12 by Lagrange. By Lemma 1, K \cong A_4, but A_4 contains no subgroups of order 6, a contradiction. Thus H \leq S_4 is maximal. \square

Now to the main results.

  1. \langle (1\ 2\ 3) \rangle and \langle (1\ 2) \rangle are distinct maximal subgroups, since each has prime index. Thus \Phi(S_3) \leq \langle (1\ 2\ 3) \rangle \cap \langle (1\ 2) \rangle = 1, so that \Phi(S_3) = 1.
  2. We know from the subgroup lattice of A_4 that \langle (1\ 2\ 3) \rangle and \langle (1\ 2\ 4) \rangle are distinct maximal subgroups of prime order. Thus \Phi(A_4) = 1.
  3. Note that H_1 = \langle (a\ b\ c), (a\ b) \rangle and H_2 = \langle (a\ b\ d), (a\ d) \rangle are distinct subgroups of order 6 in S_4, which are maximal by Lemma 2. Evidently H_1 \cap H_2 = \langle (a\ b) \rangle. Now A_4 \leq S_4 is also maximal, and (a\ b) is an odd permutation. Thus H_1 \cap H_2 \cap A_4 = 1, and we have \Phi(S_4) = 1.
  4. \Phi(A_5) \leq A_5 is normal and proper. Since A_5 is simple, \Phi(A_5) = 1.
  5. We found in a lemma to the previous exercise that \Phi(S_5) = 1.

Sym(4) is not a commutator subgroup

Prove that there is no group whose commutator subgroup is S_4.


Suppose G is a group such that G^\prime \cong S_4. We computed in §5.4 #4 that G^{\prime\prime} \cong S_4^\prime = A_4 and G^{\prime\prime\prime} \cong V_4. Note that G^\prime/G^{\prime\prime} \cong Z_2 and G^{\prime\prime}/G^{\prime\prime\prime} \cong Z_3 are cyclic; by this previous exercise, we have G^{\prime\prime} = G^{\prime\prime\prime}, a contradiction.

Thus no such group G exists.

Compute the upper and lower central series for Alt(n) and Sym(n) where n is at least 5

Find the upper and lower central series for A_n and S_n where n \geq 5.


  1. First we consider A_n.

    Recall that the higher centers of G are characteristic, hence normal. Since A_n is simple for n \geq 5, and A_n is not abelian, we have Z_0(A_n) = 1 and Z_1(A_n) = 1. Thus the upper central series of A_n, n \geq 5, is 1.

    Now A_n^0 = A_n and A_n^1 = [A_n,A_n]. Since [(a\ b\ c), (a\ d\ e)] = (a\ c\ e), A_n^1 contains all 3-cycles, and thus A_n \leq A_n^1. Hence A_n^1 = A_n, and the lower central series of A_n is A_n.

  2. Now we consider S_n.

    We have seen previously that Z(S_n) = 1 for n \geq 5. Thus Z_1(S_n) = Z_0(S_n)$, and the upper central series of S_n is 1.

    Now S_n^0 = S_n, and we saw in §5.4 #5 that S_n^1 = S_n^\prime = A_n. Now A_n = [A_n,A_n] \leq [S_n,A_n] = S_n^2 \leq A_n, so that the lower central series of S_n is S_n \geq A_n.

Compute the upper and lower central series of Sym(4) and Alt(4)

Find the upper and lower central series of S_4 and A_4.


Recall that G^\prime = [G,G] = G^1.

  1. First we consider A_4. Since (a\ b)(c\ d)(a\ b\ c) = (d\ c\ b) and (a\ b\ c)(a\ b)(c\ d) = (a\ c\ d), Z(A_4) = 1. Thus the upper central series of A_4 is 1.

    Now A_4^0 = A_4. In §5.4 #4, we found that A_4^\prime = A_4^1 = V_4. Using the computations above, we can see that [A_4,V_4] contains all products of two 2-cycles and is contained in V_4; thus [A_4,V_4] = V_4. Thus the lower central series of A_4 is A_4 \geq V_4.

  2. First we consider S_4. We have seen previously that Z(S_4) = 1; thus the upper central series of S_4 is 1.

    Now S_4^0 = S_4. We computed that S_4^\prime = S_4^1 = A_4. Now [S_4,A_4] is contained in A_4; moreover, since (a\ d)(a\ c\ b)(a\ d)(a\ b\ c) = (a\ d\ c), [S_4,A_4] contains all 3-cycles, so that A_4 \leq S_4^2. Thus the lower central series of S_4 is S_4 \geq A_4.