# Converse of congruence condition on Sylow numbers for the prime two

## 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 $0 \le r \le m - 1$.