By Frédérique Oggier

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**Extra resources for Algebraic Methods (November 11, 2011)**

**Example text**

The next transposition is (il−2 , il−1 ), thus both terms are swapped, and we now have as last 3 terms il−1 , il , il−2 . The next swap will yield il−2 , il−1 , il , il−3 and by iterating this process, we reach the last swap (i1 , i2 ), that is i2 , i3 , . . , il , i1 as we wanted to prove. The representation has a product of transpositions is not unique, for example (2, 5, 3, 6) = (2, 6)(2, 3)(2, 5) = (5, 2)(3, 5)(6, 3) = (1, 7)(2, 6)(2, 3)(2, 5)(1, 7). We can however define an invariant of a permutation, called the parity.

Then G has at least one Sylow p-subgroup. Proof. The idea of the proof is to actually exhibit a subgroup of G of order pr . For that, we need to define a clever action of G on a carefully chosen set X. Take the set X = {subsets of G of size pr } and for action that G acts on X by left multiplication. This is clearly a welldefined action. We have that |X| = pr m pr which is not divisible by p (by the previous lemma). Recall that the action of G on X induces a partition of X into orbits: X = ⊔B(S) 42 CHAPTER 1.

This easily implies that P Q is a subgroup of G, and by the previous lemma |P Q| = pr pr |P ∩ Q| implying that |P Q| is a power of p, say pc for some c which cannot be bigger than r, since |G| = pr m. Thus pr = |P | ≤ |P Q| ≤ pr so that |P Q| = pr and thus |P | = |P Q|, saying that Q is included in P . But both Q and P have same cardinality of pr , so Q = P . The third of the Sylow Theorems tells us that all Sylow p-subgroups are conjugate. 34. (3rd Sylow Theorem). Let G be a finite group of order pr m, p a prime such that p does not divide m, and r some positive integer.