Let where is odd. Prove that the number of Sylow 2-subgroups of is .
We begin with several lemmas.
Lemma 1: If , then . Proof: The homomorphism defined on the generators of by and is injective.
Lemma 2: Let be a finite group and a prime. If is a subgroup such that does not divide , then every Sylow -subgroup of is a Sylow -subgroup of . Proof: We have and , with not dividing and and ; since does not divide the index of in , . Hence any Sylow subgroup of has order , and thus is a Sylow -subgroup in .
Now to the main result.
Suppose is odd. Then and every Sylow 2-subgroup has order 2. In this case, the Sylow 2-subgroups correspond to the elements of order 2 in . There are exactly of these: elements of the form for . Thus .
Suppose now that is even; then . We proceed by induction on the width of . (Recall that the width of an integer is the number of prime factors which divide it, including multiplicity.) Let be a Sylow 2-subgroup of .
For the base case, if the width of is 0, we have . Then . In this case, is the unique Sylow 2-subgroup, so that .
For the inductive step, suppose that for any integer of width at most , the number of Sylow 2-subgroups of is . Let be an integer of width . Then for some prime and some width integer ; without loss of generality, we may assume that is the smallest prime divisor of , so that . By the induction hypothesis, has distinct Sylow 2-subgroups. By Lemma 1, . Note also that 2 does not divide , so that by Lemma 2, has at least distinct Sylow 2-subgroups.
Now by Sylow’s Theorem, divides . Thus either or . Note that there exist elements of order 2 in which are not in (the image of) ; for instance, , since otherwise we have in the image of in , a contradiction. By Sylow’s Theorem, every 2-subgroup (For instance ) is contained in some Sylow 2-subgroup. Thus there must exist at least one other Sylow 2-subgroup in ; hence .
By induction, the result is proved for all .