Tag Archives: alternating group

Exhibit a presentation of Alt(4) using two generators

Establish a finite presentation for A_4 using two generators.


We know that A_4 is generated by an element of order 2 and an element of order 3. Let \alpha = (1\ 2)(3\ 4) and \beta = (1\ 2\ 3). Evidently, these generators satisfy the relations \alpha^2 = \beta^3 = (\alpha\beta)^3 = 1. To show that these relations provide a presentation for A_4, it suffices to show that the group \langle \alpha, \beta \ |\ \alpha^2 = \beta^3 = (\alpha\beta)^3 \rangle has order at most 24. To that end, we will compute the reduced words in this group.

  • There is one word of length 0: 1.
  • There are two words of length 1: a and b. Both are reduced by definition.
  • There are at most four words of length 2: a^2, ab, ba, and b^2. Since a^2 = 1, it is not reduced. Thus 3 words of length 2 might be reduced.
  • There are at most six words of length 3: aba, ab^2, ba^2, bab, b^2a, and b^3. Since ba^2 = b and b^3 = 1, these words are not reduced. The remaining 4 words might be reduced.
  • There are at most eight words of length 4: aba^2, abab, ab^2a, ab^3, baba, bab^2, b^2a^2, and b^2ab. Since aba^2 = ab, abab = b^2a, ab^2a = bab, ab^3 = a, baba = ab^2, and b^2a^2 = b^2, these words are not reduced. The remaining two words might be reduced.
  • There are at most four words of length 5: bab^2a = b^2ab, bab^3 = ba, b^2aba = bab^2, and b^2ab^2 = aba. None of these are reduced.

Thus any group presented by \langle a,b \ |\ a^2 = b^3 = (ab)^3 = 1 \rangle has cardinality at most 12, and we have the presentation A_4 = \langle \alpha, \beta \ |\ \alpha^2 = \beta^3 = (\alpha\beta)^3 = 1 \rangle.

A group of order 12 with no subgroup of order 6 is isomorphic to Alt(4)

Show that a group of order 12 with no subgroup of order 6 is isomorphic to A_4.


Note that 12 = 2^2 \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. Let x \in P_2 have order 2 and y \in P_3 have order 3 by Cauchy; then xy has order 6, a contradiction. Now suppose 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 |N_G(P_2 \cap Q_2)| is divisible by 2^2 and another prime; thus we have P_2 \cap Q_2 normal in G. Recall that every normal group of order 2 in a finite group is central; thus we have an element x of order 2 in the center of G. Let y \in G have order 3 by Cauchy; then xy has order 6, a contradiction.

Thus n_3(G) = 4. Let P_3 \leq G be a Sylow 3-subgroup and let N = N_G(P_3). Now N has index 4 and is self-normalizing by a previous result; in particular, N is not normal in G. Now G acts on G/N by left multiplication, yielding a permutation representation G \rightarrow S_{G/N} whose kernel K is contained in N. Note that |N| = 3, so that either K = 1 or K = N. If K = N, then for all g \in G and h \in N, we have hgN = gN, so that g^{-1}hg \in N. Thus, for all g \in G, gNg^{-1} \leq N; since G is finite, this implies that N \leq G is normal, a contradiction. Thus K = 1 and in fact G \leq S_4. Since G contains 4 \cdot 2 = 8 elements of order 3, it contains all elements of order 3 in S_4. The subgroup generated by the three cycles in S_4 is A_4, so that A_4 \leq G; since these subgroups are finite and have the same cardinality, in fact G \cong A_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.

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.

Exhibit a representative of each cycle type in Alt(5) as a commutator in Sym(5)

Exhibit a representative of each cycle type in A_5 as a commutator in S_5.


The cycle types in A_5 are (5), (3), and (2,2).

  1. (a\ b)(c\ d)(a\ c\ e)(a\ b)(c\ d)(a\ e\ c) = (a\ b\ d\ e\ c)
  2. (a\ c\ b)(a\ b)(a\ b\ c)(a\ b) = (a\ b\ c)
  3. (a\ b)(c\ d)(a\ c\ b)(a\ b)(c\ d)(a\ b\ c) = (a\ d)(b\ c)

For n at least 5, the commutator subgroup of Sym(n) is Alt(n)

Prove that the commutator subgroup of S_n is A_n for all n \geq 5.


We know that [S_n,S_n] is characteristic, hence normal, in S_n. Since n \geq 5, then, this commutator is either 1, A_n, or S_n.

Now [S_n,S_n] \neq 1 since S_n is nonabelian. Moreover, since S_n/A_n \cong Z_2 is abelian, we have that [S_n, S_n] \leq A_n. Thus [S_n,S_n] = A_n.

Compute the commutator subgroups of Sym(4) and Alt(4)

Find the commutator subgroups of S_4 and A_4.


We start with S_4. Note that A_4 is normal in S_4 and S_4/A_4 \cong Z_2 is abelian, so that by Proposition 7, S_4^\prime \leq A_4. Note that (a\ c\ b)(a\ b)(a\ b\ c)(a\ b) = (a\ b\ c); that is, (a\ b\ c) = [(a\ b\ c), (a\ b)]. Now S_4^\prime is a characteristic, hence normal, subgroup, and thus is a union of conjugacy classes. Since the conjugacy classes in S_4 are precisely the cycle shapes, all 3-cycles are in S_4^\prime.

Since (a\ b)(c\ d)(a\ c\ b)(a\ b)(c\ d)(a\ b\ c) = (a\ d)(b\ c), all products of two 2-cycles are also in S_4^\prime. With the identity, this exhausts the elements of A_4; hence S_4^\prime = A_4.

Now we consider A_4. We know that V_4 \leq A_4 is characteristic, hence normal, and that (by Lagrange) A_4/V_4 \cong Z_3 is abelian. By Proposition 7, A_4^\prime \leq V_4. Note moreover that since (a\ b\ c)(a\ b)(c\ d)(a\ c\ b) = (a\ d)(b\ c), every order 2 subgroup is not normal in A_4, hence not characteristic, and thus cannot be the commutator subgroup. Thus A_4^\prime = V_4.

Construct an infinite simple group

Let \Omega = \{1,2,3,\ldots\} (as in this previous exercise). Let D be the subgroup of S_\Omega consisting of permutations which move only a finite number of elements of \Omega (described in this previous exercise) and let A be the set of all elements \sigma \in D that act as even permutations on the (finite) set M(\sigma). Prove that A is an infinite simple group. [Hint: show that every pair of elements of A lie in a finite simple subgroup of A.]


We begin with a lemma.

Lemma: If x,y \in A, then there is a finite simple group N \leq A with x,y \in N. Proof: If x,y \in A, then M(x) and M(y) are finite. Let S = \{a_1,a_2,a_3,a_4,a_5\} be five elements in \Omega not moved by either x or y. Now T = M(x) \cup M(y) \cup S is finite, and by Theorem 24 in the text, A_T \leq A is simple. Clearly x,y \in A_T. \square

Now suppose H \leq A is a normal subgroup. Suppose H \neq 1; let x \in H be a nonidentity. Suppose that there exists y \in A \setminus H. Now there exists a finite simple group K \leq A with x,y \in K. Moreover, H \cap K \leq K is a nontrivial normal subgroup (containing in particular x \neq 1) so that, since K is simple, H \cap K = K. Thus K \leq H. But then y \in H, a contradiction. Thus no such element y exists, and we have H = A.

Thus if H \leq A is a normal subgroup, either H = 1 or H = A. Hence A is simple.

Finally, it is clear that A is infinite since, for example, A_n \leq A is a subgroup for every integer n.

Alt(n) is generated by the set of all 3-cycles

Prove that A_n is generated by the set of all 3-cycles for n \geq 3.


Let n \geq 3 and let T = \{ (a\ b\ c) \ |\ 1 \leq a,b,c \leq n \} be the set of 3-cycles in A_n.

Note that A_n contains T, so that \langle T \rangle \leq A_n.

Recall that A_n consists of all permutations which can be written as an even product of transpositions; more specifically, A_n is generated by the set of all products of two distinct 2-cycles. (Here we use the fact that n \geq 3.) Each product \sigma of two 2-cycles has one of two forms.

If \sigma = (a\ b)(c\ d), note that \sigma = (a\ c\ b)(a\ c\ d).

If \sigma = (a\ b)(a\ c), note that \sigma = (a\ c\ b).

Thus A_n \leq \langle T \rangle, hence A_n = \langle T \rangle.