After comparing $H/(H\cap K)$ with $HK/K$, we now study what happens when one quotient is taken after another. The focus keyword third isomorphism theorem describes the rule $$\begin{equation*} (G/H_1)/(H_2/H_1)\cong G/H_2, \end{equation*}$$ where $H_1\subseteq H_2$ are normal subgroups of $G$. The theorem says that first collapsing $H_1$ and then collapsing $H_2/H_1$ has the same effect as collapsing $H_2$ in one step. Students often find quotient-of-quotient notation intimidating, but the idea is only staged simplification. :::theorem[Third Isomorphism Theorem] Let $H_1\trianglelefteq G$ and $H_2\trianglelefteq G$ with $H_1\subseteq H_2$. Then $H_2/H_1\trianglelefteq G/H_1$, and $$\begin{equation*} (G/H_1)/(H_2/H_1)\cong G/H_2. \end{equation*}$$ ::: :::proof Given that $H_1\trianglelefteq G$, $H_2\trianglelefteq G$, and $H_1\subseteq H_2$. To prove that $H_2/H_1\trianglelefteq G/H_1$ and $(G/H_1)/(H_2/H_1)\cong G/H_2$. Define $\phi:G/H_1\to G/H_2$ by $$\begin{equation*} \phi(aH_1)=aH_2. \end{equation*}$$ [1] To prove that $\phi$ is well-defined. Suppose $aH_1=bH_1$. Then $b^{-1}a\in H_1$. Since $H_1\subseteq H_2$, we have $b^{-1}a\in H_2$. Therefore $aH_2=bH_2$. Thus $\phi(aH_1)=\phi(bH_1)$. [2] To prove that $\phi$ is a homomorphism. Let $aH_1,bH_1\in G/H_1$. Then $$\begin{align*} \phi((aH_1)(bH_1))&=\phi(abH_1)\\ &=abH_2,\\ \phi(aH_1)\phi(bH_1)&=(aH_2)(bH_2)\\ &=abH_2. \end{align*}$$ Thus $\phi((aH_1)(bH_1))=\phi(aH_1)\phi(bH_1)$. [3] To prove that $\phi$ is onto. Every element of $G/H_2$ has the form $aH_2$, and $$\begin{equation*} aH_2=\phi(aH_1). \end{equation*}$$ [4] To compute the kernel. $$\begin{align*} \ker\phi&=\{aH_1\in G/H_1\mid \phi(aH_1)=H_2\}\\ &=\{aH_1\in G/H_1\mid aH_2=H_2\}\\ &=\{aH_1\mid a\in H_2\}\\ &=H_2/H_1. \end{align*}$$ Since kernels are normal subgroups, $H_2/H_1\trianglelefteq G/H_1$. By the first isomorphism theorem, $$\begin{equation*} (G/H_1)/(H_2/H_1)\cong G/H_2. \end{equation*}$$ $\square$ ::: The quotient $G/H_1$ has already made every element of $H_1$ into the identity coset. Inside this quotient, the subgroup $H_2/H_1$ represents what remains of $H_2$. Collapsing that remaining subgroup finishes the job of collapsing all of $H_2$. Thus the third isomorphism theorem tells us that quotienting can be performed in stages without changing the final structure. This is especially useful when large quotient groups are easier to understand through smaller intermediate quotients. :::example[Quotients of the Integers] In $(\mathbb{Z},+)$, let $H_1=(6)$ and $H_2=(3)$. Since $(6)\subseteq(3)$ and every subgroup of $\mathbb{Z}$ is normal, the third isomorphism theorem gives $$\begin{equation*} (\mathbb{Z}/(6))/((3)/(6))\cong\mathbb{Z}/(3). \end{equation*}$$ Thus a quotient of $\mathbb{Z}/(6)$ by the subgroup $(3)/(6)$ is isomorphic to $\mathbb{Z}_3$. ::: :::example[Quotienting by Stages] Let $N\subseteq H\subseteq G$ with $N\trianglelefteq G$ and $H\trianglelefteq G$. Then the third isomorphism theorem gives $$\begin{equation*} (G/N)/(H/N)\cong G/H. \end{equation*}$$ The first quotient collapses $N$. The second quotient collapses the image of $H$ inside $G/N$. Together, these two steps collapse exactly $H$ in the original group. ::: This preview uses subgroups $(m)\subseteq(n)$ of $\mathbb Z$, so the divisibility condition is $n\mid m$. First $\mathbb Z$ is collapsed modulo $m$, and then the subgroup $(n)/(m)$ is collapsed inside $\mathbb Z/(m)$. Adjust $m$ and $n$ and observe the size of the intermediate quotient, the subgroup being collapsed, and the final quotient. Try $m=12,n=4$ and then enter a pair where $n$ does not divide $m$ to see why containment is required. :::scientific-preview[Quotient Stage Explorer] ::: The middle group remembers residues modulo $m$. The subgroup $(n)/(m)$ is what remains of the larger subgroup after the first collapse. Quotienting by that remaining part gives the same final structure as reducing modulo $n$ from the start. This calculator verifies the integer version of the third isomorphism theorem. Enter $m$ and $n$ with $n\mid m$, corresponding to $(m)\subseteq(n)$. The calculator reports the size of the intermediate group, the size of the subgroup being collapsed, and the size of the final quotient. Use it to confirm that quotienting by stages reduces the size from $m$ to $n$ exactly as $\mathbb Z/(n)$ does. :::calculator[Quotient Stage Calculator] ::: The number of cosets after the second quotient is exactly the order of the larger quotient $\mathbb Z/(n)$. The theorem strengthens this size match into an isomorphism of groups. :::solved-problem Let $(m)\subseteq(n)\subseteq\mathbb{Z}$. Use the third isomorphism theorem to write a quotient of $\mathbb{Z}/(m)$ isomorphic to $\mathbb{Z}/(n)$. ::: :::solution Let $(m)\subseteq(n)\subseteq\mathbb{Z}$. Then $n\mid m$. Take $H_1=(m)$ and $H_2=(n)$. Since $\mathbb{Z}$ is abelian, both subgroups are normal in $\mathbb{Z}$. By the third isomorphism theorem, $$\begin{equation*} (\mathbb{Z}/(m))/((n)/(m))\cong\mathbb{Z}/(n). \end{equation*}$$ Thus the quotient of $\mathbb{Z}/(m)$ by the subgroup $(n)/(m)$ is isomorphic to $\mathbb{Z}/(n)$. $\square$ ::: :::solved-problem Let $G$ be a group and let $H_1\subseteq H_2$ be normal subgroups of $G$. Identify the natural epimorphism from $G/H_1$ onto $G/H_2$ and its kernel. ::: :::solution Let $H_1\subseteq H_2$ be normal subgroups of $G$. Define $\phi:G/H_1\to G/H_2$ by $$\begin{equation*} \phi(aH_1)=aH_2. \end{equation*}$$ This is well-defined because $aH_1=bH_1$ implies $b^{-1}a\in H_1\subseteq H_2$, so $aH_2=bH_2$. The map is onto because every coset $aH_2$ is the image of $aH_1$. Its kernel is $$\begin{align*} \ker\phi&=\{aH_1\mid aH_2=H_2\}\\ &=\{aH_1\mid a\in H_2\}\\ &=H_2/H_1. \end{align*}$$ $\square$ ::: :::tip In the third isomorphism theorem, check the containment direction carefully. The smaller normal subgroup $H_1$ must be inside the larger normal subgroup $H_2$, so that $H_2/H_1$ is a subgroup of $G/H_1$. ::: :::exercise 1. Apply the theorem to $(12)\subseteq(4)\subseteq\mathbb{Z}$. 2. If $H_1=H_2$, what does the theorem say? 3. If $H_1=\{e\}$, what does the theorem reduce to? ::: :::answer 1. The theorem gives $(\mathbb{Z}/(12))/((4)/(12))\cong\mathbb{Z}/(4)$. 2. It gives $(G/H_1)/(H_1/H_1)\cong G/H_1$, quotienting by the trivial subgroup. 3. It gives $(G/\{e\})/(H_2/\{e\})\cong G/H_2$, which is essentially $G/H_2$. ::: :::faq[Frequently Asked Questions] Q: Why is $H_2/H_1$ normal in $G/H_1$? A: It is the kernel of the natural epimorphism $G/H_1\to G/H_2$. Q: What does the theorem say conceptually? A: Quotienting by stages gives the same result as quotienting directly by the larger normal subgroup. Q: Can the theorem be used for non-normal subgroups? A: No. The quotient groups involved require normality. ::: :::call-to-action[Continue to the Correspondence Theorem] subtitle: Learn how subgroups above a kernel correspond exactly to subgroups of the epimorphic image or quotient group. button: Study the Correspondence Theorem | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/isomorphism-and-correspondence-theorems/correspondence-theorem button-ghost: Review the Second Isomorphism Theorem | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/isomorphism-and-correspondence-theorems/second-isomorphism-theorem :::