How are there two C3 rotation axes in ammonia? If we look at the character table for the $C_\mathrm{3v}$ point group (which $\ce{NH3}$ is an example of), we see that these are the symmetry elements ${E, 2 C_3, 3 \sigma_v}$ present in the molecule. My question is — why are there $2 C_3$ rotation axes? Shouldn't there be only one, which is perpendicular to the plane containing the three hydrogens?
The $C_\mathrm{3v}$ character table:
$$\begin{array}{c|ccc|cc} \hline
C_\mathrm{3v} &amp; E &amp; 2C_3 &amp; 3\sigma_\mathrm{v} &amp; &amp; \\ \hline
\mathrm{A_1} &amp; 1 &amp; 1 &amp; 1 &amp; z &amp; x^2+y^2, z^2 \\ 
\mathrm{A_2} &amp; 1 &amp; 1 &amp; -1 &amp; R_z &amp;  \\ 
\mathrm{E} &amp; 2 &amp; -1 &amp; 0 &amp; (x,y), (R_x,R_y) &amp; (x^2-y^2,xy), (xz,yz) \\ \hline
\end{array}$$
In a character table for $C_{3v}$, $ 2C_3$, means that the element is actually 2 $C_3$ elements in the same conjugacy class.  That means the two $C_3$ operations are related in some way.
In fact, it means they share the same single axis.  One $C_3$ operation is a $120^\circ$ turn around the axis, and the other $C_3$ operation is a $240^\circ$ turn around the same axis in the same direction - or equivalently a $120^\circ$ turn around the same axis in the opposite direction.
In the image below the two elements $C_3$ and $C_3^{-1}$ are the two symmetry elements of the $2C_3$ class, which are rotations about the same axis out of the plain of the image.

Symmetry elements of the $C_{3v}$ point group
Image taken from Fig S5 in supplementary information of T. Karin, et al, Phys. Rev. B 94, 041201