Suppose $F$ is a nonabelian free group and $F/N_1$ and $F/N_2$ are amenable groups. Does it follow that $F/(N_1 \cap N_2)$ is amenable?
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$F/N_1\cap N_2$ embeds into the product $F/N_1 \times F/N_2$ under the natural map. Since the class of amenable groups is closed under products and passing to subgroups, we see that $F/N_1\cap N_2$ is indeed amenable.