No, because there are not enough of them. The lowest Steenrod operation $\mathcal{P}^1$ raises degree by $2(q-1) = 4 \tfrac{q-1}{2}$, so $\mathcal{P}^1(p_1)$ has degree $4(\tfrac{1-1}{2}+1)$ and so you need at least $p_1, p_2, \ldots, p_{\tfrac{1-1}{2}}$ in a generating set.

No, because there are not enough of them. The lowest Steenrod operation $\mathcal{P}^1$ raises degree by $2(q-1) = 4 \tfrac{q-1}{2}$, so $\mathcal{P}^1(p_1)$ has degree $4(\tfrac{q-1}{2}+1)$ and so you need at least $p_1, p_2, \ldots, p_{\tfrac{q-1}{2}}$ in a generating set. ${}$${}$