9.4. MORE ON ALGEBRAIC FIELD EXTENSIONS 229

Proposition 9.4.36 Let n≥ 3. Then every permutation in An is the product of 3 cycles andthe identity.

Proof: In case n = 3, you can list all of the permutations in An.(1 2 31 2 3

),

(1 2 32 3 1

),

(1 2 33 1 2

)In terms of cycles, these are

identity, (1,2,3) ,(1,3,2)

You can easily check that the the last two are inverses of each other.Now suppose n ≥ 4. The permutations in An are defined as the product of an even

number of transpositions. There are two cases. The first case is where you have twotranspositions which share a number,

(a,c)(c,b) = (a,c,b)

Thus when they share a number, the product is just a 3 cycle. Next suppose you have theproduct of two transpositions which are disjoint. This can happen because n≥ 4. First notethat

(a,b) = (c,b)(b,a,c) = (c,b,a)(c,a)

Therefore,

(a,b)(c,d) = (c,b,a)(c,a)(a,d)(d,c,a)

= (c,b,a)(c,a,d)(d,c,a)

and so every product of disjoint transpositions is the product of 3 cycles. ■

Lemma 9.4.37 If n ≥ 5, then if B is a normal subgroup of An, and B is not the identity,then B must contain a 3 cycle.

Proof: Let α be the permutation in B which is “closest” to the identity without beingthe identity. That is, out of all permutations which are not the identity, this is one whichhas the most fixed points or equivalently moves the fewest numbers. Then α is the productof disjoint cycles. Suppose that the longest cycle is the first one and it has at least fournumbers. Thus

α = (i1, i2, i3, i4, · · · ,m)γ1 · · ·γ p

Since B is normal,

α1 ≡ (i3, i2, i1)(i1, i2, i3, i4, · · · ,m)(i1, i2, i3)γ1 · · ·γ p ∈ Am

Then since the various cycles are disjoint, α1α−1 =

(i3, i2, i1)(i1, i2, i3, i4, · · · ,m)(i1, i2, i3)γ1

· · ·γ p (m, · · · , i4, i3, i2, i1)γ−1p · · ·γ−1

1

= (i3, i2, i1)(i1, i2, i3, i4, · · · ,m)(i1, i2, i3)(m, · · · , i4, i3, i2, i1)γ1

· · ·γ pγ−1p · · ·γ−1

1

= (i3, i2, i1)(i1, i2, i3, i4, · · · ,m)(i1, i2, i3)(m, · · · , i4, i3, i2, i1)