By Miller G. A.

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**Example text**

Gn of subgroups of G, where G0 = {1} and Gn = G, such that Gi−1 is normal in Gi and Gi /Gi−1 is Abelian for i = 1, 2, . . , n. Example The symmetric group Σ4 is solvable. Indeed let V4 be the Kleinsche Viergruppe consisting of the identity permutation ι and the permutations (12)(34), (13)(24) and (14)(23), and let A4 be the alternating group consisting of all even permutations of {1, 2, 3, 4}. Then {ι} V4 A4 Σ4 , V4 is Abelian, A4 /V4 is cyclic of order 3, and Σ4 /A4 is cyclic of order 2. 54 Let G be a group, let H1 and H2 be subgroups of G, where H1 H2 , and let J1 = H1 ∩ N , J2 = H2 ∩ N , K1 = H1 N/N and K2 = H2 N/N , where N is some normal subgroup of G.

Now the number np of Sylow p-subgroups divides pq and satisfies np ≡ 1 (mod p), by the Second Sylow Theorem. Clearly np cannot be divisible by p, and therefore either np = 1 or np = q. But q ≡ 1 (mod p). It follows that np = 1. Thus the group G has just one subgroup of order p. Now, given any element g of G, the subgroups Np and gNp g −1 are of order p. It follows that gNp g −1 = Np for all elements g of G. Thus Np is a normal subgroup of G. A similar argument shows that Nq is also a normal subgroup of G, since p < q, and therefore p ≡ 1 (mod q).

Otherwise the group G contains at least two distinct subgroups H and K of order 9. 52 that |H ∩ K| ≥ 94 . But |H ∩ K| divides 9, by Lagrange’s Theorem, since H ∩ K is a subgroup of H and of K. Therefore |H ∩ K| = 3. 53 we see that H ∩ K is a normal subgroup of H and of K. Let J = {g ∈ G : g(H ∩ K)g −1 = H ∩ K}. Then J is a subgroup of G, and H ∩ K is a normal subgroup of J. Moreover H and K are subgroups of J, and therefore |J| is divisible by 9, by Lagrange’s Theorem. But J is a subgroup of G, and hence |J| divides 36.