Let and be finite groups and let be a prime. Prove that any Sylow -subgroup of has the form , where and are Sylow -subgroups of and , respectively. Deduce that . Generalize these results to a direct product of finitely many finite groups (so that the number of Sylow -subgroups of a product is the product of the numbers of Sylow -subgroups of the factors).
Let and be finite groups and a prime, and write and , where does not divide either or . Note that , and that does not divide .
Suppose and are Sylow -subgroups. Then and , so that . Thus is a Sylow -subgroup.
Now suppose is a Sylow -subgroup.
Define and . We claim that and are subgroups of and , respectively; we prove this for only. Suppose ; then for some . Then , so that . By the Subgroup Criterion, . Similarly .
Note that if , then for some . We also have , so that and have -power order. Thus and are -subgroups, as otherwise some nonidentity element does not have -power order. By Sylow’s Theorem, there exist Sylow -subgroups and with and .
Note that , but that . Because these sets are finite, .
Thus we have proved that ; hence .
The generalization to arbitrary finite direct products proceeds by induction as follows.
Suppose that for some , for an arbitrary direct product of groups , every Sylow -subgroup of is a product of Sylow -subgroups of the (and vice versa). Let be arbitrary. Then every Sylow -subgroup of is of the form where and are Sylow -subgroups (and vice versa); by the induction hypothesis, for Sylow -subgroups . Thus every Sylow -subgroup of has the form for some Sylow -subgroups (and vice versa).
Thus we have for all groups and positive integers .