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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?

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    $\begingroup$ 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 $H_2$... $\endgroup$ – Yu Ding Apr 22 at 3:56
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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.

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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 $H_1\cap H_2N\supseteq H_2$, …

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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 $H_{1} \cap H_{2} N$.)

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