number theory question in a group with unknown order I was reading a paper and I am struggling understanding one part of it.
Lets say we have a group $G$ of an unknown order $n$. we know that $B&lt;n&lt;B+C$. both B and C are large values).

we choose a random element $z \in G$. and let's say we have some prime values $B+C&lt;e_1,e_2,...,e_l$. We Calculate $h=z^ {\prod_i e_i}$.

The claim is as $\gcd(n,\prod e_i)=1$ (which is true as each prime is greater than $n$) we can say that $h$ is uniformly distributed in $G$. How can we prove this statement?
Define the function
$$\begin{align}
f: G&amp;\to G\\
x&amp;\mapsto x^{\prod_i e_i}
\end{align}$$
$f$ is injective. Proof: if $f(x)=f(y)$ then $x^{\prod_i e_i}=y^{\prod_i e_i}$, thus $(x\cdot y^{-1})^{\prod_i e_i}=1$, thus the order $k$ of $(x\cdot y^{-1})$ is a divisor of $\prod_i e_i$. Since the order of any group element divides the group order, $k$ is also a divisor of $n$. Since $\gcd(n,\prod e_i)=1$, $k$ must be $1$. Thus $(x\cdot y^{-1})=1$, thus $x=y$.

Any injection over a finite set is a bijection. It follows that $f$ is a bijection over $G$.

Thus $h$ is constructed as the image of a uniformly random element $z$ of $G$ by a bijection $f$ over $G$. Hence $h$ is a uniformly random element of $G$.