# Read e-book online Algebra [Lecture notes] PDF

By I. M. Isaacs

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As examples of how to apply the previous theorem, we do two examples. 1. If |G| = 120 then G is not simple. Proof. As |G| = 120 = 23 · 3 · 5, we know that n5 ∈ {1, 2, 4, 8, 3, 6, 12, 24}. Yet n5 ≡ 1 mod 5, so n5 ∈ {1, 6}. If we wish for G to be simple, we assume that n5 = 6. So let H = NG (P ), where P ∈ Syl5 (G). So |G : H| = 6 by our previous conclusion. Let G act via right multiplication on Ω = {Hx|x ∈ G}. Now θ : G → Sym(Ω) via the “dotting” map and θ is a homomorphism. As |Ω| = 6, we see that Sym(Ω) ∼ = S6 , so θ may be thought of as a map from G to S6 .

X1 y1 y2 . . yk−1 where the result, xk yk−1 lies in Nk as xk , yk ∈ Nk . However, due to the right hand side of the equation above, the result is also in N1 n2 . . Nk−1 . By our hypothesis, we see that we must then have that xk yk−1 = 1, and hence xk = yk . Yet this is a contradiction to the fact that k is our minimal criminal, so we must have xk = yk for all 1 ≤ i ≤ r. 2. Let G be finite and nilpotent and let p1 , p2 , . . pr be the distinct prime divisors of |G|. Let Pi be the unique element of the set of Sylow pi subgroups.

Let g ∈ Ni ∩ Nj . Then we may write g = t1 t2 . r where ti = g and i=j tj = 1 for all j = i, as g ∈ Ni . Also, as g ∈ Nj , we may write g = s1 s2 . . sr j=i where sk ∈ Nk for all k and si = 1. By the “directness” of the product, we have that sk = tk for all k, so g = ti = si = 1. 34 We draw as a corollary the following important fact. 1. If G is the internal direct product of Ni for 1 ≤ i ≤ r then Ni ∩ Nj = 1 for i = j and so the elements of Ni commute with the elements of Nj . We now proceed with the proof of the theorem.