Let $G \leq SL(V)$ be a finite group where $V$ is a finite dimensional linear space over a finite field $k$, and assume that every element in $G$ has a nonzero fixed point in $V$, then is it necessary that there exist a nonzero point fixed by $G$ ?
I think in the case $dim V=2$, this is true. For example, if $G \leq SL_2(k)$, then one consider the fixed line $l_g$ for any $g \not =1 \in G$, then $l_g$ must be the same or one choose two lines as a basis then get a contradiction by considering the product.
No. For instance, suppose $\operatorname{char} k\neq 2$, let $V=k^3$, and let $G\subseteq SL(V)$ be the subgroup generated by $$\begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{pmatrix}$$ and $$\begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{pmatrix}.$$ Explicitly, an element of $G$ is an even permutation matrix except that two of its $1$s may be changed to $-1$. Every element of $G$ fixes some nonzero vector: if the permutation is the identity then some standard basis vector is fixed, if the permutation is a 3-cycle and there are no $-1$s then $(1,1,1)$ is fixed, and if the permutation is a 3-cycle and there are two $-1$s then some permutation of the vector $(1,1,-1)$ is fixed. However, no nonzero vector is fixed by $G$ (the first generator only fixes scalar multiples of $(1,1,1)$, which are not fixed by the second generator).