Let be a group with more than one Sylow -subgroup. Over all pairs of distinct Sylow -subgroups let and be chosen so that is maximal. Show that has more than one Sylow -subgroup and that any two distinct Sylow -subgroups of intersect in . (Thus is divisible by and by some prime other than . Note that Sylow -subgroups of need not by Sylow in .)
Note that is proper since is a -group. Since , we have that divides .
Suppose now that is a -group. Then for some Sylow -group . If , then , and hence . However, since and , we have a contradiction since there are elements in and but not in (namely, ). Thus . But then we have and , so that . This implies that is not (cardinality) maximal among the pairwise intersections of Sylow -subgroups. Thus is not a -group, so that its order is divisible by some prime other than .
Suppose now that contains a unique Sylow -subgroup, say . Then . Moreover, for some Sylow -sugbroup of . Since and is order maximal, we in fact have . Similarly, . But then , so that is Sylow in , a contradiction.
Let . Since is a -group, is contained in some Sylow -group . By Sylow’s Theorem, is conjugate to and in , say by and , respectively. Since is normal in , . Thus . Now and for some Sylow -subgroups and of . Suppose . Then, since and are distinct, the subgroup as generated in is a -group, normalizes , and properly contains (for instance) , which is a contradiction since is Sylow in . Thus we have . Moreover, , and since is maximal among the intersections of distinct Sylow -subgroups and is finite, .