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Math Help - normal Sylow subgroups

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    normal Sylow subgroups

    Let G be a finite group with |G|=pqr, where p, q, r are primes with p<q<r. Prove that G has a normal Sylow subgroup for either p, q, or r.
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    Quote Originally Posted by dori1123 View Post

    Let G be a finite group with |G|=pqr, where p, q, r are primes with p<q<r. Prove that G has a normal Sylow subgroup for either p, q, or r.
    suppose to the contrary that neither of Sylow subgroups of G are normal. let n_p, n_q, n_r be the number of Sylow p, q and r subgroups of G respectively. so n_p > 1, \ n_q > 1, n_r > 1. now n_r \mid pqr,

    which gives us n_r \mid pq because \gcd(n_r,r)=1. but since n_r > r > q > p, we have n_r \neq p, \ n_r \neq q. thus n_r=pq. now since n_q \mid pr, and n_q > q > p, we have n_q = r, \text{or} \ pr. in either case n_q \geq r.

    similarly n_p \geq q. so we've proved that G has exactly pq subgroups of order r, at least r subgroups of order q and at least q subgroups of order p. since every two distinct cyclic subgroups of the

    same prime order have trivial intersection and also every two cyclic subgroups of distinct prime orders have trivial intersection, a simple counting of elements of G shows that we must have:

    |G| \geq pq(r-1) + (q-1)r + (p-1)q + 1=pqr + (q-1)r - q + 1 > pqr =|G|, which is the contradiction we needed! Q.E.D.
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