If I understand correctly, binary variable $g_j$ indicates whether group $j$ is special, binary variable $a_{i,j}$ indicates whether exactly $i$ items are taken from group $j$, and binary variable $b_i$ indicates whether exactly $i$ items are taken from the special group. You want to enforce the logical implication
$$b_i \implies \bigvee_j (g_j \land a_{i,j})$$
One approach is to introduce a new binary variable $c_{i,j}$ and enforce
\begin{align}
c_{i,j} &\implies (g_j \land a_{i,j}) \tag1 \\
b_i &\implies \bigvee_j c_{i,j} \tag2
\end{align}
Rewriting $(1)$ in conjunctive normal form yields
$$
c_{i,j} \implies (g_j \land a_{i,j}) \\
\lnot c_{i,j} \lor (g_j \land a_{i,j}) \\
(\lnot c_{i,j} \lor g_j) \land (\lnot c_{i,j} \lor a_{i,j}) \\
(1 - c_{i,j} + g_j \ge 1) \land (1 - c_{i,j} + a_{i,j} \ge 1) \\
(c_{i,j} \le g_j) \land (c_{i,j} \le a_{i,j})
$$
Rewriting $(2)$ in conjunctive normal form yields
$$
b_i \implies \bigvee_j c_{i,j} \\
\lnot b_i \lor \bigvee_j c_{i,j} \\
1 - b_i + \sum_j c_{i,j} \ge 1 \\
b_i \le \sum_j c_{i,j}
$$
So the desired linear constraints are
\begin{align}
c_{i,j} &\le g_j \\
c_{i,j} &\le a_{i,j} \\
b_i &\le \sum_j c_{i,j}
\end{align}
From your comment, it sounds like you also want to enforce the converse of $(1)$:
$$(g_j \land a_{i,j}) \implies c_{i,j}$$
One approach is to again use conjunctive normal form to obtain
$$g_j + a_{i,j} - 1 \le c_{i,j} \quad \text{for all $i$ and $j$} $$
But because there is only one count per group, you have $\sum_i a_{i,j}=1$ and can instead use "compact linearization" to obtain fewer constraints. Explicitly, summing both sides of
$$c_{i,j} = g_j a_{i,j}$$ over $i$ yields
$$\sum_i c_{i,j} = g_j \tag3$$
Because there is only one special group, you have $\sum_j g_j=1$ and can further sum over $j$ to obtain just one constraint:
$$\sum_{i,j} c_{i,j} = 1 \tag4$$
as you suggested.
Constraint $(3)$ yields a tighter LP formulation because constraint $(4)$ is an aggregation, but either one is sufficient to enforce your desired behavior.
