By Frédérique Oggier

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

**Sample text**

THE JORDAN-HOLDER THEOREM 55 Since G/H1 is simple, we get that G1 /(H1 ∩ G1 ) is simple as well. Now by the above lemma, upon removing duplicates from the series {1} = H1 ∩ Gn ··· H1 ∩ G0 = H1 , we get a composition series for H1 of length at most n and thus upon removing duplicates {1} = H1 ∩ Gn · · · H1 ∩ G1 is a composition series for H1 ∩ G1 of length at most n − 1. Since G1 /(H1 ∩ G1 ) is simple, it follows that upon removing duplicates {1} = H1 ∩ Gn ··· H1 ∩ G1 ⊳ G1 is a composition series for G1 .

Let gP ∈ Y be the element whose orbit size is 1. We have h · gP = gP for h ∈ R, since gP belongs to its orbit. Thus g −1 hg ∈ P ⇐⇒ h ∈ gP g −1 for all h in R. We have just proved that the p-group R is contained in a conjugate of P . All we needed for the proof is that R is a p-group, so the same proof holds for the case of a Sylow p-subgroup, for which we get that R is contained in a conjugate of P , and both have same cardinality, which concludes the proof. We will use the fact that the proof works for R a p-group in general for proving one corollary.

If G is the direct product of its Sylow subgroups, that every Sylow subgroup of G is normal is immediate since the factors of a direct product are normal subgroups. 37, we know that every normal Sylow p-subgroup is unique, thus there is a unique Sylow pi -subgroup Pi for each prime divisor pi of |G|, i = 1, . . , k. 32, we have that |P1 P2 | = |P1 ||P2 | since P1 ∩ P2 = {1}, and thus |P1 · · · Pk | = |P1 | · · · |Pk | = |G| by definition of Sylow subgroups. Since we work with finite groups, we deduce that G is indeed the direct product of its Sylow subgroups, having that G = P1 · · · Pk and Pi ∩ j=i Pj is trivial.

### Algebraic Methods (November 11, 2011) by Frédérique Oggier

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