Is there a way to confirm that a homomorphic division (multiplication with inverse) using ElGamal produced the correct result? By running some tests I observed that if I perform a homomorphic division (multiplication with the multiplicative inverse) between two values using the ElGamal scheme, I get the correct result when the division produces a whole number.
For example, D(E(10) / E(2)) = 5 which is the expected result (here &quot;/&quot; denotes division applied homomorphically).
But if I try D(E(10) / E(3)) the result is a very large positive or negative number.
This might be expected since, I think, ElGamal supports homomorphic division only when the dividend is divisible by the divisor. My question is given only the decrypted result of such operation, is there a way to know whether the division &quot;succeeded&quot; in the sense that it is the result of a division where the dividend is divisible by the divisor?
I've included some discussion in general, but &quot;El Gamal&quot; here is a red herring. Your confusion is with the difference between &quot;division&quot; in $\mathbb{R}$, and &quot;division&quot; in $\mathbb{F}_p$. I discuss this more in detail at the end.

El Gamal's ciphertexts are of the form:
$$\mathsf{Enc}_{g, h}(m) = (g^y, mh^y)$$
Note that $h = g^x$, where $x$ is the secret key.
This admits a (multiplicative) homomorphism, as you mention:
$$\mathsf{Enc}_{g, h}(m_1)\ast \mathsf{Enc}_{g, h}(m_2) = (g^{y_1+y_2}, m_1m_2 h^{y_1+y_2})$$
I believe your confusion stems from confusion over what $m_1m_2h^{y_1+y_2}$ means.
Like many cryptosystems, El Gamal is defined over what is known as a &quot;finite field&quot; (in particular $\mathbb{Z}/p\mathbb{Z}\cong\mathbb{F}_p$). This is a number system that allows you to add, subtract, multiply, and divide (just like with $\mathbb{R}$), but the operations are somehwat different than you are used to. Fix $p = 5$ (although any prime number works). Then addition is defined &quot;mod 5&quot;, so:
$$3 + 3\bmod 5 \equiv 6\bmod 5\equiv 1\bmod 5$$
So in this number system, $3 + 3 = 1$.
Multiplication is similarly defined &quot;mod 5&quot;.
So:
$$3 \times 3\bmod 5 \equiv 9\bmod 5 \equiv 4\bmod 5$$
So $3\times 3 = 4$ in this number system.
What does &quot;division&quot; mean in this number system?
Well, one can show that for any (non-zero) $x\in\mathbb{Z}/p\mathbb{Z}$, there is a unique $y$ such that $xy \equiv 1\bmod p$.
Over $\mathbb{R}$ this is true as well (instead with the equation $xy = 1$) --- if $x = 2$, then $y = 1/2$.
But in $\mathbb{Z}/5\mathbb{Z}$, if we have that $x = 2$, then $y = 3$ means that $xy\bmod 5\equiv 6\bmod 5\equiv 1\bmod 5$.
So, in $\mathbb{Z}/5\mathbb{Z}$, we have that $2^{-1} = 1/2\equiv 3\bmod p$.
This is now when we get to your point --- if we have $x,y\in\mathbb{Z}/p\mathbb{Z}$, then $xy/y\equiv xyy^{-1}\equiv x\bmod p$, so &quot;when $y$ divides $xy$, division is like you would expect over $\mathbb{R}$&quot;.
But the situation can be more complicated in general.