Does there exist such a subgroup of a linear group? Could anybody give an example of the following case : $F$ is a field with nonzero characteristic $p$, $V$ is a $F$-vector space with finite dimension, $G$ is a subgroup of $GL(V)$, $G$ is infinite but has finite exponent and its (least) exponent is not divisible by $p$ ? Thanks in advance.

This does not exist. With no loss of generality, you can suppose that the field is algebraically closed. Let $H$ be the Zariski closure of $G$ and $H_0$ its unit component. Then $H$ is a connected positive-dimensional algebraic group, of finite exponent. Since tori have infinite exponent, $H_0$ has to be unipotent. Hence its exponent has the form $p^n$ with $n\ge 1$. Hence $p$ divides the exponent of $G$.

Added: To avoid the torus argument (but still with a little smaller amount of Zariski topology), one observes that since elements of $H$ have bounded order, they achieve finitely many characteristic polynomials, and hence by connectedness it only achieves on $H_0$ the characteristic polynomial $(X-1)^k$ where $k$ is the ambient dimension. So all elements of $H_0$ are unipotent, and hence their order is a power of $p$. Thus the exponent of $H_0$ is $p^m$ for some $m\ge 0$, and actually $m>0$ since $H_0\neq 1$.
