Can I perform a division of two integers homomorphically using ElGamal? How can I perform a division of two integers homomorphically? 
(Simplifying assumptions can be made if needed, that is, I am fine with dividing numbers that are whole and the result will be whole as well, e.g., simple divisions such as 4 divided by 2 are fine)

Given two ElGamal ciphertexts:
$$ C_1 = (C^{1}_{1}, C^2_1) $$
$$ C_2 = (C^{1}_{2}, C^2_2) $$

I can multiply them homomorphically as follows:
$$ C_r = (C^{1}_{1} \cdot C^{1}_{2} ,C^2_1 \cdot C^2_2) $$

The following does not seem to work to achieve division though:
$$ C_r = (C^{1}_{1} \div C^{1}_{2} ,C^2_1 \div C^2_2) $$
Let $G$ is a cyclic group with order $q$ and generator $g$.

Let $x$ be Alice's private key. The two messages $c_1$ and $c_2$ are encrypted with the public key of Alice;

$$c_1 = (g^{y_1}, m_1 \cdot g^{x y_1})$$

$$c_2 = (g^{y_2}, m_2 \cdot g^{x y_2})$$

Now to calculate $m_1/m_2$ in the encrypted values, calculate the inverse of $g^{y_2}$ and $m_2 \cdot g^{x y_2}$ in the group

$$ c_2'=\Big((g^{y_2})^{-1},(m_2 \cdot g^{x y_2} )^{-1}\Big) = (g^{-y_2},m_2^{-1} \cdot g^{-x y_2})$$

then multiply, 

$$c_1  c_2' = (g^{y_1}  g^{-y_2}, \;\; m_1 m_2^{-1} \cdot g^{x y_1} g^{- x y_2})$$

Now, check the decryption.

$$s = (g^{y_1} g^{-y_2})^x = g^{x y_1 }  g^{-x y_2 } $$

$$s^{-1} =  g^{-x y_1 } \cdot g^{x y_2 } $$

$$ s^{-1} \cdot m_1 m_2^{-1} \cdot g^{x y_1} g^{- x y_2} =  m_1  m_2^{-1}  \cdot g^{-x y_1 }g^{x y_1}\cdot  g^{x y_2 } g^{- x y_2} = m_1  m_2^{-1}$$



Note: based on comments of Geoffroy and Mikeazo.