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 $0\leq r\leq m-1$.