11. The Alternating Groups Consider the group S3 . Then this group contains a normal subgroup, generated by a 3-cycle. Now the elements of S3 come in three types. The identity, the prod uct of zero transpositions, the transpositions, the product of one trans position, and the three cycles, products of two transpositions. Then the normal subgroup above, consists of all permutations that can be represented as a product of an even number of transpositions. In general there is no canonical way to represent a permutation as a product of transpositions. But we might hope that the pattern above continues to hold in every permutation group. Deﬁnition 11.1. Let σ ∈ Sn be a permutation. We say that σ is even if it can be represented as a product of an even number of transpositions. We say that σ is odd if it can be represented as a product of an odd number of transpositions. The following result is much trickier to prove than it looks. Lemma 11.2. Let σ ∈ Sn be a permutation. Then σ is not both an even and an odd permutation. There is no entirely satisfactory proof of (11.2). Here is perhaps the simplest. Deﬁnition 11.3. Let x1 , x2 , . . . , xn be indeterminates and set (xi − xj ). f (x1 , x2 , . . . , xn ) = i b then xa − xi is sent xb − xi and there is no change in sign. Similarly if i = a and j 6= b. If j < a or j > b there is no change in sign. If a < j < b there is a change in sign. But then the ﬁrst two cases contribute in total an even number of signs changes (in fact, there will be exactly (a − b − 1) + (a − b − 1) = 2(a − b − 1) sign changes). Finally we need to consider the case i = a and j = b. In this case xa − xb gets replaced by xb − xa and there is a change in sign. Hence (3). D Proof. Suppose that σ is a product of an even number of transpositions. Then by (2) and (3) of (11.5), σ ∗ (f ) = f . Similarly if σ ∗ (f ) is a product of an odd number of transpositions, then σ ∗ (f ) = −f . Thus σ cannot be both even and odd. D Deﬁnition-Lemma 11.6. There is a surjective homomorphism φ : Sn −→ Z2 The kernel consists of the even transpositions, and is called the alter nating group An . Proof. The map sends an even transposition to 1 and an odd transposi tion to −1. (2) of (11.5) implies that this map is a homomorphism. D 2 MIT OCW: 18.703 Modern Algebra Prof. James McKernan Note that half of the elements of Sn are even, so that the alternating group An contains n2! . One of the most important properties of the alternating group is, 6 4. Theorem 11.7. Suppose that n = The only normal subgroup of Sn is An . Moreover An is simple, that is, An has no proper normal subgroups. If n = 4 then we have already seen that {e, (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}. is a normal subgroup of S4 . In fact it is also a normal subgroup of A4 , so that A4 is not simple. 3 MIT OCW: 18.703 Modern Algebra Prof. James McKernan

The Alternating Groups

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