# Suppose that $H_1/H_2$ is Abelian. Show that $H_1 N / H_2 N$ is Abelian.

Suppose $$G$$ is a group and $$H_1$$, $$H_2$$, $$N$$ are subgroups of $$G$$. $$N$$ is a normal subgroup of $$G$$ and $$H_2$$ is a normal subgroup of $$H_1$$. Suppose that $$H_1/H_2$$ is Abelian. Show that $$H_1N/H_2N$$ is Abelian.

My current thought is using third isomorphism theorem and deduct $$H_1N/H_2N$$ is a quotient of $$H_1/H_2$$. However, I just cannot prove $$H_1N/H_2N$$ is a quotient of $$H_1/H_2$$. Does my thought wrong?

• Consider the map $H_1\to H_1N/H_2N$ defined by $h\mapsto \langle h e\rangle$. This is a surjective homomorphism whose kernel contains $H_2$... – Yu Ding Apr 22 at 3:56

This can be proven using only the definition of the quotient. Suppose $$H_1/H_2$$ is abelian, then for any $$a,b \in H_1$$ we have $$aba^{-1}b^{-1} =h \in H_2$$. Now let $$an,bm\in H_1N$$ be given. We have \begin{align} anbmn^{-1}a^{-1}m^{-1}b^{-1} & = an(a^{-1}a)bmn^{-1}a^{-1}(b^{-1}b)m^{-1}b^{-1}\\ & = n'abmn^{-1}a^{-1}b^{-1}m'\\ & = n'abmn^{-1}[(ab)^{-1}(ab)]a^{-1}b^{-1}m'\\ & = n'm''aba^{-1}b^{-1}m'\\ & = n'm''hm' \end{align} where $$n',m',m'' \in N$$ and their existence follows from normality of $$N$$. As $$n'm''hm'\in H_2N$$ by definition, we have $$anbm(bman)^{-1} \in H_2N$$ hence $$H_1N/H_2N$$ is abelian.
Your idea is good. After you have proved that $$H_2N$$ is a normal subgroup of $$H_1N$$, you can consider the map $$H_1\to H_1N/H_2N,\qquad x\mapsto xH_2N$$ Prove this map is a surjective homomorphism. The kernel is $$H_1\cap H_2N$$. Since $$H_1\cap H_2N\supseteq H_2$$, …
A proof using the second and third isomorphism theorems goes like this $$\frac{H_{1} N}{H_{2} N} = \frac{H_{1} H_{2} N}{H_{2} N} \cong \frac{H_{1}}{H_{1} \cap H_{2} N} \cong \frac{H_{1} / H_{2}}{(H_{1} \cap H_{2} N) / H_{2}},$$ and the latter is abelian, as an homomorphic image of the abelian group $$H_{1}/H_{2}$$. (Note that $$H_{2}$$ is indeed contained in $$H_{1} \cap H_{2} N$$.)