Converse of congruence condition on Sylow numbers for the prime two

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Statement

For any odd number m, there exists a finite group G where the number of 2-Sylow subgroup (?)s of G (i.e., the Sylow number (?) for the prime 2) equals m.

Proof

Further information: Dihedral group, subgroup structure of dihedral group

We take G as the dihedral group D_{2m} of degree m and order 2m. It is given by the presentation:

G := \langle a,x \mid a^m = x^2 = e, xax = a^{-1} \rangle.

The two-element subgroup \langle x \rangle is a 2-Sylow subgroup, and it is a self-normalizing subgroup. It has m conjugates, given by \langle a^r x \rangle, with 0 \le r \le m - 1.