How much do you know about groups structure?

The square has 90 degree rotational symmetry and if you repeatedly apply that same rotation, you'll reach the square's original orientation after exactly 4 rotations.

Therefore the answer is yes, either element corresponding to a 90 degree rotation has order 4.

Solution:

We have $G = D_4 =$

$\left\{\underbrace{\left(\begin{array}{cccc} 1 & 2 & 3 & 4\\ 1 & 2 & 3 & 4 \end{array} \right)}_{Id}, \underbrace{\left(\begin{array}{cccc} 1 & 2 & 3 & 4\\ 2 & 3 & 4 & 1 \end{array} \right)}_{\varphi}, \underbrace{\left(\begin{array}{cccc} 1 & 2 & 3 & 4\\ 3 & 4 & 1 & 2 \end{array} \right)}_{\varphi^2}, \underbrace{\left(\begin{array}{cccc} 1 & 2 & 3 & 4\\ 4 & 1 & 2 & 3 \end{array} \right)}_{\varphi^3}, \underbrace{\left(\begin{array}{cccc} 1 & 2 & 3 & 4\\ 1 & 4 & 3 & 2 \end{array} \right)}_{\psi}, \underbrace{\left(\begin{array}{cccc} 1 & 2 & 3 & 4\\ 2 & 1 & 4 & 3 \end{array} \right)}_{\varphi \cdot \psi} , \underbrace{\left(\begin{array}{cccc} 1 & 2 & 3 & 4\\ 3 & 2 & 1 & 4 \end{array} \right)}_{\varphi^2 \cdot \psi}, \underbrace{\left(\begin{array}{cccc} 1 & 2 & 3 & 4\\ 4 & 3 & 2 & 1 \end{array} \right)}_{\varphi^3 \cdot \psi} \right\}$

$\Rightarrow \varphi^4 = \varphi \cdot \varphi^3 = \left(\begin{array}{cccc} 1 & 2 & 3 & 4\\ 2 & 3 & 4 & 1 \end{array} \right) \cdot \left(\begin{array}{cccc} 1 & 2 & 3 & 4\\ 4 & 1 & 2 & 3 \end{array} \right) = \left(\begin{array}{cccc} 1 & 2 & 3 & 4\\ 1 & 2 & 3 & 4 \end{array} \right) = Id \Rightarrow |\varphi| = 4$

answer: exists