25 Beautiful Homes - March 2012 by Editor: Deborah Barker

By Editor: Deborah Barker

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Proof. We first prove that ψ is a bijection. Injectivity. If H1 /N = H2 /N , then cosets in each subgroup are the same, that is for any h1 ∈ H1 , we have h1 N = h2 N for some h2 ∈ H2 , implying that h−1 2 h1 ∈ N ⊂ H2 and thus h1 ∈ H2 , showing that H1 ⊆ H2 . By repeating the same argument but reverting the role of H1 and H2 , we get H2 ⊆ H1 and thus H1 = H2 . Surjectivity. Let Q be a subgroup of G/N and let π : G → G/N be the canonical projection. Then π −1 (Q) = {a ∈ G, aN ∈ Q}. This is a subgroup of G containing N and ψ(π −1 (Q)) = {aN, aN ∈ Q} = Q.

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.

Thanks to the two above results, we can now prove that a non-abelian simple group must have more than one Sylow p-subgroup. 38. Let G be a finite group which is non-abelian and simple. If the prime p divides |G|, then the number np of Sylow p-subgroups is strictly bigger than 1. Proof. Let us look at the prime factors appearing in the order of G. • If p is the only prime factor of |G|, then |G| must be a power of p, that is G is a non-trivial p-group (it is non-trivial by definition of simple and a p-group by a corollary of the 1st Sylow Theorem).

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