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

**Example text**

G · Q = gQg −1 , g ∈ P , Q ∈ X. Note that in the case where P is the only Sylow p-subgroup, then we can take Q = P . By the Orbit-Stabilizer Theorem, we have that the orbit B(Q) of any Sylow p-subgroup Q in X has cardinality |B(Q)| = |P |/|Stab(Q)| = pr /|Stab(Q)|. 8. THE SYLOW THEOREMS In particular, the size of an orbit of any Sylow p-subgroups divides pr , meaning it has to be either 1 or a power of p. Let us recall that the set X is partitioned by the orbits B(Q) under the action of P , so that the cardinality of X is: |X| = |B(Q)| = |B(Q′ )| + |B(Q′′ )| where Q′ and Q′′ denote subgroups whose orbit has respectively one element or at least two elements.

Since Z(G) is a normal subgroup of G and G is simple, it must be that G = Z(G), which contradicts the assumption that G is non-abelian. • We then know that |G| is divisible by at least two distinct primes. So if P is a Sylow p-subgroup, then {1} < P < G, where the second inclusion is strict since the order of G is divisible by two primes. If there were only one Sylow p-subgroup, namely np = 1, then this Sylow p-subgroup would be normal by the above lemma, which contradicts the simplicity of G. Let us see if we can be more precise by refining the assumptions on the order of the group G we consider.

Suppose that Gi+1 is not a maximal normal subgroup of Gi , then we can refine the subnormal series by inserting a group H such that Gi+1 ⊳ H ⊳ Gi , and we can repeat this process hoping it will terminate (it will if G is finite, it may not otherwise). 31. Let G be a group, and let G0 , . . , Gn be subgroups of G such that 1. Gn = {1} and G0 = G, 2. Gi+1 ⊳ Gi , i = 0, . . , n − 1, such that Gi+1 is a maximal normal subgroup of Gi . Then the series {1} = Gn ⊳ Gn−1 ⊳ · · · ⊳ G0 = G is called a composition series for G.