# normal Sylow subgroups

• Nov 10th 2008, 01:20 PM
dori1123
normal Sylow subgroups
Let G be a finite group with $|G|=pqr$, where $p, q, r$ are primes with $p. Prove that G has a normal Sylow subgroup for either p, q, or r.
• Nov 10th 2008, 08:42 PM
NonCommAlg
Quote:

Originally Posted by dori1123

Let G be a finite group with $|G|=pqr$, where $p, q, r$ are primes with $p. 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.