10.3. Simplicity of the Alternating Groups
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10. SOLVABILITY
In the following exercises, you can use whatever criterion for solvabil ity appears most convenient.
10.2.5. Show that any subgroup of a solvable group is solvable.
10.2.6. Show that any quotient group of a solvable group is solvable.
10.2.7. Show that if N G is a normal subgroup and both N and G=N
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are solvable, then also G is solvable.
10.3. Simplicity of the Alternating Groups
In this section, we will prove that the symmetric groups Sn and the alternating groups An for n 5 are not solvable. In fact, we will see that the alternating group An is the unique nontrivial proper normal subgroup of Sn for n 5, and moreover that An is a simple group for n 5.
Recall that conjugacy classes in the symmetric group Sn are deter mined by cycle structure. Two elements are conjugate precisely when they have the same cycle structure. This fact is used frequently in the following.
Lemma 10.3.1. For n 3, the alternating group An is generated by 3– cycles.
Proof. A product of two 2–cycles in Sn is conjugate to .12/.12/ D e D .123/.132/, if the two 2–cycles are equal; or to .12/.23/ D .123/, if the two 2–cycles have one digit in common; or to .12/.34/ D .132/.134/, if the two 2–cycles have no digits in common. Thus, any product of two 2–cycles can be written as a product of one or two 3–cycles. Any even permutation is a product of an even number of 2–cycles and, therefore, can be written as a product of 3–cycles.
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Since the center of a group is always a normal subgroup, if we want to show that a nonabelian group is simple, it makes sense to check first that the center is trivial. You are asked to show in Exercise 10.3.1 that for n 3 the center of Sn is trivial and in Exercise 10.3.4 that for n 4 the center of An is trivial.
Theorem 10.3.2. If n 5 and N is a normal subgroup of Sn such that N ¤ feg, then N An.
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Proof. By Exercise 10.3.1, the center of Sn consists of e alone. Let ¤ e be an element of N ; since is not central and since 2–cycles generate Sn, there is some 2–cycle such that ¤ . Consider the element 1; writing this as . / 1 and using the normality of N , we see that this element is in N ; on the other hand, writing the element as . 1/, we see that the element is a product of two unequal 2–cycles.
If these two 2–cycles have a digit in common, then the product is a 3–cycle.
If the two 2–cycles have no digit in common, then by normality of N , N contains all elements that are a product of two disjoint 2–cycles. In particular, N contains the elements .12/.34/ and .12/.35/ and, therefore, the product .12/.34/.12/.35/ D .34/.35/ D .435/. So also in this case, N contains a 3–cycle.
By normality of N , N contains all 3–cycles, and, therefore, by Lemma 10.3.1, N An.
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Lemma 10.3.3. If n 5, then all 3–cycles are conjugate in An.
Proof. It suffices to show that any 3–cycle is conjugate in An to .123/. Let be a 3–cycle. Since and .123/ have the same cycle structure, there is an element of 2 Sn such that .123/ 1 D . If is even, there is nothing more to do. Otherwise, 0 D .45/ is even, and 0.123/0 1 D .
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Theorem 10.3.4. If n 5, then An is simple.
Proof. Suppose N ¤ feg is a normal subgroup of An and that ¤ e is an element of N . Since the center of An is trivial and An is generated by 3–cycles, there is a 3–cycle that does not commute with . Then (as in the proof of the previous theorem) the element 1 1 is a nonidentity element of N that is a product of two 3–cycles.
The rest of the proof consists of showing that N must contain a 3– cycle. Then by Lemma 10.3.3, N must contain all 3–cycles, and, therefore, N D An by Lemma 10.3.1.
The product of two 3–cycles must be of one of the following types:
1. .a1a2a3/.a4a5a6/
2. .a1a2a3/.a1a4a5/ D .a1a4a5a2a3/
3. .a1a2a3/.a1a2a4/ D .a1a3/.a2a4/
4. .a1a2a3/.a2a1a4/ D .a1a4a3/
5. .a1a2a3/.a1a2a3/ D .a1a3a2/
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10. SOLVABILITY
In either of the last two cases, N contains a 3–cycle.
In case (3), since n 5, An contains an element .a2a4a5/. Compute that .a2a4a5/.a2a4a5/ 1 D .a1a3/.a4a5/, and the product 1.a2a4a5/.a2a4a5/ 1 D .a2a5a4/ is a 3–cycle in N .
In case (2), compute that .a1a4a3/.a1a4a3/ 1 D .a4a3a5a2a1/, and 1.a1a4a3/.a1a4a3/ 1 D .a4a2a3/ is a 3–cycle in N .
Finally, in case (1), compute that .a1a2a4/.a1a2a4/ 1 D .a2a4a3/.a1a5a6/; and the product 1.a1a2a4/.a1a2a4/ 1 is .a1a4a6a2a3/, namely, a 5–cycle. But then by case (2), N contains a 3–cycle.
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These computations may look mysterious and unmotivated. The idea is to take a 3–cycle x and to form the commutator 1xx 1. This is a way to get a lot of new elements of N . If we experiment just a little, we can find a 3–cycle x such that the commutator is either a 3–cycle or, at the worst, has one of the cycle structures already dealt with.
Here is an alternative way to finish the proof, which avoids the com putations. I learned this method from I. Herstein, Abstract Algebra, 3rd edition, Prentice Hall, 1996.
I claim that the simplicity of An for n > 6 follows from the simplicity of A6. Suppose n > 6 and N ¤ feg is a normal subgroup of An. By the first part of the proof, N contains a nonidentity element that is a product of two 3–cycles. These two 3–cycles involve at most 6 of the digits 1 k n; so there is an isomorphic copy of S6 in Sn such that N \ A6 ¤ feg.
Since N \A6 is a normal subgroup of A6, and A6 is supposed to be simple, it follows that N \ A6 D A6, and, in particular, N contains a 3–cycle. But if N contains a 3–cycle, then N D An, by an argument used in the original proof.
So it suffices to prove that A5 and A6 are simple. Take n D 5 or 6, and suppose that N ¤ feg is a normal subgroup of An that is minimal among all such subgroups; that is, if feg K N is a normal subgroup of An,
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then K D feg.
Let X be the set of conjugates of N in Sn, and let Sn act on X by conjugation. What is the stabilizer (centralizer) of N ? Since N is normal in An, CentS .N / A
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n. There are no subgroups properly between An and Sn, so the centralizer is either An or Sn. If the centralizer is all of Sn, then N is normal in Sn, so by Theorem 10.3.2, N D An.
If the centralizer of N is An, then X has 2 D ŒSn W An elements. Let M ¤ N be the other conjugate of N in Sn. Then M D N 1 for any i
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odd permutation . Note that M Š N and, in particular, M and N have the same number of elements.
I leave it as an exercise to show that M is also a normal subgroup of An.
Since N ¤ M , N \M is a normal subgroup of An properly contained in N . By the assumption of minimality of N , it follows that N \M D feg.
Therefore, MN is a subgroup of An, isomorphic to M N . The group MN is normal in An, but if is an odd permutation, then MN 1 D M 1N 1 D NM D MN , so MN is normal in Sn. Therefore, by Theorem 10.3.2 again, MN D An. In particular, the cardinality of An is jM N j D jN j2. But neither 5Š=2 D 60 nor 6Š=2 D 360 is a perfect square, so this is impossible.
This completes the alternative proof of Theorem It follows from the simplicity of An for n 5 that neither Sn nor An is solvable for n 5 (Exercise 10.3.6).
Exercises 10.3 10.3.1. Show that for n 3, the center of Sn is feg. Hint: Suppose that 2 Z.An/. Then the conjugacy class of consists of alone. What is the cycle structure of ?
10.3.2. Show that the subgroup V D fe; .12/.34/; .13/.24/; .14/.23/g of A4 is normal in S4.
10.3.3. Show that if A is a normal subgroup of a group G, then the center Z.A/ of A is normal in G.
10.3.4. This exercise shows that the center of An is trivial for all n 4.
(a) Compute that the center of A4 is feg.
(b) Show that for n 4, An is not abelian.
(c) Use Exercise 10.3.3 and Theorem 10.3.2 to show that if n 5, then the center of An is either feg or An. Since An is not abelian, the center is feg.
10.3.5. Show that the subgroup M appearing in the alternative proof of 10.3.4 is a normal subgroup of An.
10.3.6. Observe that a simple nonabelian group is not solvable, and conclude that neither An nor Sn are solvable for n 5.
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