Suppose that $ \Gamma \subset F^*_p $ (integers modulo a prime number $p$) is a multiplicative subgroup. $ A \subset F^*_p$ is a set and $ A $ is smaller than $ \Gamma $. Also
$$ A\Gamma = \{a\gamma: a \in A, \gamma \in \Gamma \} $$
is a set of products of every element of $ A $ by every element of $\Gamma$.
Is it true that $|A\Gamma| = |A|$?
The answer is no. An easy example is to take $|A|=1$, because multiplication in $F^*$ is cancellative; then $|A\Gamma|=|\Gamma|$, which by assumption is strictly larger than $A$. Thus, for instance, in $\mathbb{F}_5$ you can take $\Gamma=\{1,-1\}$ and $A=\{2\}$.