# Why is the El Gamal commitment scheme information theoretically binding?

I am a bit stuck on the following claim:

The ElGamal commitment scheme is information theoretically binding

As far as I understand, an adversary $A$ would win the binding game if it is able to find an $x\prime\neq x$ such that $C(x)=C(x\prime)$.

$C(x)=C(x\prime)$ implies:

$$(g^r,h^rx)=(g^{r\prime},h^{r\prime}x\prime)$$

How is this impossible to be found? Since generators are cyclic, it should be possible to find an $r\neq r\prime$ that with $g^r=g^{r\prime}$, or am I overseeing something?

How is this impossible to be found? Since generators are cyclic, it should be possible to find an $r\neq r'$ that with $g^r = g^{r'}$, or am I overseeing something?
Yes, you're probably thinking the wrong way, possibly like this: there are values $a,b$ with $a\neq b$ and $a^2 = b^2$. But if you look closely, in this case the base is the same and the exponent differs.
The main arguement here is: If the group order is $t$, and we have $r \neq r' \mod t$, then $g^{r} \neq g^{r'}$. Alternatively: set $x = r' - r \mod t$, which is non-zero modulo $t$. Then $g^{r'} = g^{r + x} = g^r \cdot g^x$. This can't be equal to $g^r$ if $x$ is not zero or a multiple of the group order.
With a fixed generator, the first element in the tuple fixes $r$ - even if you don't know what $r$ actually is. And that in return fixes $h^r$. And for a given ElGamal ciphertext this fixes $m$.