Could I prove this result in probability theory when the random variables are defined in fields/groups or abelian groups?

Suppose that $l$, $y$ and $x$ are three different random variables that are uniformly distributed. Say that $l$ is defined over a finite field (or a group) $G$, $y\in Y$ such that $|Y|\geq |G|$ and define $\rho:G\times Y\to X$ such that $\phi(\cdot,y_i)$ is bijective, namely the pair $(l,y)\in G\times Y$ is associated with exactly one $x_i$. As it hilds in general for group theory every field or group or abelian group is endowed with the operations of summation $\oplus_{G}$ ($\ominus_{G}$ resp. the subtraction) in $G$ and $\otimes_{G}$ in G. Can we prove the following?

If $l$ is a random variable with support on $G$ (since it takes vales on it) and $y$ is uniformly distributed over $G$ and independent of $l$, then the random variable defined as $x=l\ominus_{G}y$ is also uniformly distributed over $G$.

Solution 1:

I will use multiplicative notation to allow $G$ to be non-abelian. Let

• $G$ be any finite group,
• $L$ be a random variable taking values in $G$,
• $Y$ be a random variable that is uniformly distributed over $G$ and is independent of $L$.

Then $X=LY^{-1}$ is also uniformly distributed over $G$, since for each $g\in G$,

$$ \mathbf{P}(X=g) = \mathbf{P}(Y=g^{-1}L) = \sum_{l\in G}\mathbf{P}(Y = g^{-1}l)\mathbf{P}(L=l) = \sum_{l\in G} \frac{1}{|G|}\mathbf{P}(L=l) = \frac{1}{|G|}.$$