Having seen that the kernel records exactly which elements a homomorphism sends to the identity, we now ask when a homomorphism can be rebuilt from a quotient group. The focus keyword factorization through quotients means that a homomorphism $f:G\to G_1$ may pass through $G/H$ whenever $H$ is contained in the kernel of $f$. In that situation, all elements of the same coset of $H$ have the same image under $f$. The quotient has already collapsed $H$, so the new map from $G/H$ to $G_1$ is well-defined. This lesson prepares the first isomorphism theorem by explaining the exact mechanism behind quotient factorization. :::definition[Factorization Through a Quotient] Let $f:G\to G_1$ be a homomorphism and let $H\trianglelefteq G$. We say that $f$ factors through the quotient $G/H$ if there exists a homomorphism $\overline{f}:G/H\to G_1$ such that $$\begin{equation*} f=\overline{f}\circ \pi, \end{equation*}$$ where $\pi:G\to G/H$ is the natural homomorphism defined by $\pi(a)=aH$. ::: The quotient group $G/H$ treats all elements of a coset $aH$ as one object. If a map from $G/H$ is defined by $\overline{f}(aH)=f(a)$, then this value must not depend on the chosen representative $a$. If $aH=bH$, then $b^{-1}a\in H$. Thus $f(a)=f(b)$ is guaranteed when every element of $H$ is sent to the identity. This is exactly the condition $H\subseteq\ker f$. A common mistake is to assume normality alone is enough; normality makes the quotient a group, while the kernel condition makes the induced map well-defined. :::theorem[Factorization Through Quotients] Let $f:G\to G_1$ be a homomorphism, and let $H\trianglelefteq G$ satisfy $H\subseteq\ker f$. Let $\pi:G\to G/H$ be the natural homomorphism. Then there exists a unique homomorphism $\overline{f}:G/H\to G_1$ such that $$\begin{equation*} f=\overline{f}\circ\pi. \end{equation*}$$ It is defined by $\overline{f}(aH)=f(a)$. Moreover, $\overline{f}$ is one-to-one if and only if $H=\ker f$. ::: :::proof Given that $f:G\to G_1$ is a homomorphism, $H\trianglelefteq G$, $H\subseteq\ker f$, and $\pi(a)=aH$. To prove that there exists a unique homomorphism $\overline{f}:G/H\to G_1$ such that $f=\overline{f}\circ\pi$, and that $\overline{f}$ is one-to-one if and only if $H=\ker f$. Define $\overline{f}:G/H\to G_1$ by $\overline{f}(aH)=f(a)$. [1] To prove that $\overline{f}$ is well-defined. Suppose $aH=bH$. Then $b^{-1}a\in H$. Since $H\subseteq\ker f$, we have $f(b^{-1}a)=e_1$. Hence $$\begin{align*} f(b)^{-1}f(a)&=e_1,\\ f(a)&=f(b). \end{align*}$$ Therefore $\overline{f}(aH)=\overline{f}(bH)$. [2] To prove that $\overline{f}$ is a homomorphism. Let $aH,bH\in G/H$. Then $$\begin{align*} \overline{f}((aH)(bH))&=\overline{f}(abH)\\ &=f(ab)\\ &=f(a)f(b)\\ &=\overline{f}(aH)\overline{f}(bH). \end{align*}$$ Thus $\overline{f}$ is a homomorphism. [3] For every $a\in G$, $$\begin{equation*} (\overline{f}\circ\pi)(a)=\overline{f}(aH)=f(a). \end{equation*}$$ Therefore $f=\overline{f}\circ\pi$. [4] To prove uniqueness, let $u:G/H\to G_1$ be a homomorphism such that $f=u\circ\pi$. For every $aH\in G/H$, $$\begin{equation*} u(aH)=u(\pi(a))=(u\circ\pi)(a)=f(a)=\overline{f}(aH). \end{equation*}$$ Hence $u=\overline{f}$. [5] The kernel of $\overline{f}$ is $$\begin{align*} \ker\overline{f}&=\{aH\in G/H\mid \overline{f}(aH)=e_1\}\\ &=\{aH\in G/H\mid f(a)=e_1\}\\ &=(\ker f)/H. \end{align*}$$ Therefore $\overline{f}$ is one-to-one if and only if $\ker\overline{f}=\{H\}$, which occurs if and only if $\ker f=H$. Hence the theorem is proved. $\square$ ::: :::example[Remainder Map Through a Larger Quotient] Let $f:(\mathbb{Z},+)\to(\mathbb{Z}_3,+_3)$ be defined by $f(n)=[n]_3$, and let $H=(6)$. Since $\ker f=(3)$ and $(6)\subseteq(3)$, the map $f$ factors through $\mathbb{Z}/(6)$. The induced map is $$\begin{equation*} \overline{f}(n+(6))=[n]_3. \end{equation*}$$ If $n+(6)=m+(6)$, then $6\mid n-m$, so $3\mid n-m$, and hence $[n]_3=[m]_3$. Thus the formula is well-defined. ::: This preview tests the proposed rule from a quotient group of integers to a remainder group. Choose a modulus $m$ for the map $f(n)=[n]_m$, choose a quotient subgroup $(h)$, and choose a representative $r$. Compare $r$ with $r+h$, because these two integers represent the same coset modulo $h$. We observe a well-defined induced map exactly when both representatives always land in the same remainder class modulo $m$. Try the default values $m=3$ and $h=6$, then change $h$ to $4$ to see why the kernel condition matters. :::scientific-preview[Coset Representative Test] ::: The green case means the quotient has already collapsed only elements that $f$ sends to the identity. The red case means one coset would receive two different values, so the proposed induced map is not a function. This is the concrete role of the condition $(h)\subseteq\ker f$. This calculator checks the integer model $f:\mathbb{Z}\to\mathbb{Z}_m$ with $f(n)=[n]_m$ and asks whether it factors through $\mathbb{Z}/(h)$. Enter positive integers $m$ and $h$. The calculator reports whether $(h)\subseteq(m)$, whether the induced map exists, and whether the induced map is one-to-one. Use it to compare $h=m$, where the induced map is injective, with larger multiples of $m$, where extra kernel remains. :::calculator[Quotient Factorization Checker]