If a group has order G 385 show that its 11Sylow subgroup i

If a group has order |G| = 385 show that its 11-Sylow subgroup is normal in G and its 7-Sylow subgroup lies in the center of G.

Solution

Every group of order 385 contains a central Sylow 7-subgroup and a normal Sylow 11-subgroup. Indeed, note that 385=5711385=5711. Now Sylow’s Theorem forces n7=n11=1, so that G has a unique, hence normal Sylow 7- and a normal Sylow 11-subgroup. Let P7 denote the (unique) Sylow 7-subgroup.

We have NG(P7)=G. Hence we obtain G/CG(P7)Aut(P7).

Moreover, |Aut(P7)|=6=23. Thus |G/CG(P7)|=1, so that G/CG(P7)=1, hence CG(P7)=G. ThusP7Z(G).