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Suppose that 𝐻,𝑁,𝑀 are subgroups of 𝐺, 𝑀 is a normal subgroup of 𝑁. Assume 𝑁/𝑀 is Abelian. Use the second isomorphism theorem to show that (𝑁∩𝐻)/(𝑀∩𝐻) is Abelian.

I can show (𝑁∩𝐻)M is a subgroup of 𝑁. However, I don't know how to proceed from here.

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The second isomorphism theorem says that, given a subgroup $S$ of a group $G$ and a normal subgroup $N$ of $G,$ then

  • $SN$ is a subgroup of $G,$
  • $S\cap N$ is a normal subgroup of $S,$ and
  • $(SN)/N\cong S/(S\cap N).$

Here, though, our (sub)groups are named differently. Instead of $S,$ we're considering $N\cap H;$ instead of $N,$ $M;$ instead of $G,$ $N.$

You're right that $(N\cap H)M$ is a subgroup of $N$ (the first point above), but we also should observe that $$(N\cap H)\cap M=(N\cap M)\cap H=M\cap H$$ is a normal subgroup of $N$ (the second point above, using the fact that $M\subseteq N$) and that $(N\cap H)M/M\cong (N\cap H)/(M\cap H)$ (the third point above).

I would then show that $(N\cap H)M/M$ is a subgroup of the abelian group $N/M,$ so is abelian, itself, whence the proof is complete.

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