# Use the second isomorphism theorem to show that (𝑁∩𝐻)/(𝑀∩𝐻) is Abelian.

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.

• – Shaun Apr 21 at 21:14

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.