First, we need to show that $g(H \cap K) \subseteq gH \cap gK$.
Let $x \in g(H \cap K)$. Then, there exists an element $h \in H \cap K$ such that $x = gh$. Since $h \in H \cap K$, we have $h \in H$ and $h \in K$. Thus, $gh \in gH$ and $gh \in gK$. Therefore, $x
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