After constructing quotient groups, we next ask which familiar properties survive after passing from $G$ to $G/H$. Two important answers are immediate but powerful: a quotient of an abelian group is abelian, and a quotient of a cyclic group is cyclic. These results allow us to recognise many quotient groups without writing a complete Cayley table. The key idea is that each coset $aH$ remembers the behaviour of the representative $a$, but only up to multiplication by elements of $H$. Students often confuse the statement with a converse; a quotient may be abelian even when the original group is not. :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $G$ is abelian, then $G/H$ is abelian. ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $G$ is abelian. To prove that $G/H$ is abelian. Since $G$ is abelian, every subgroup of $G$ is normal in $G$. Therefore $G/H$ is a quotient group. Let $aH,bH\in G/H$. Since $G$ is abelian, $$\begin{equation*} a\circ b=b\circ a. \end{equation*}$$ Therefore $$\begin{align*} aH*bH &=(a\circ b)H\\ &=(b\circ a)H\\ &=bH*aH. \end{align*}$$ Thus $$\begin{equation*} aH*bH=bH*aH \quad \forall aH,bH\in G/H. \end{equation*}$$ Hence, $G/H$ is abelian. $\square$ ::: :::example Let $G=(\mathbb{Z},+)$ and let $H=4\mathbb{Z}$. Since $\mathbb{Z}$ is abelian, the quotient group $\mathbb{Z}/4\mathbb{Z}$ is abelian. Its elements are $$\begin{equation*} 0+4\mathbb{Z},\quad 1+4\mathbb{Z},\quad 2+4\mathbb{Z},\quad 3+4\mathbb{Z}. \end{equation*}$$ For example, $$\begin{equation*} (1+4\mathbb{Z})+(3+4\mathbb{Z})=0+4\mathbb{Z}. \end{equation*}$$ Hence, $\mathbb{Z}/4\mathbb{Z}$ is an abelian quotient group. $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $G$ is cyclic, then $G/H$ is cyclic. ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $G$ is cyclic. To prove that $G/H$ is cyclic. Since every cyclic group is abelian, $G$ is abelian. Therefore $H$ is normal in $G$, and $G/H$ is a quotient group. Since $G$ is cyclic, there exists $a\in G$ such that $$\begin{equation*} G=\langle a\rangle. \end{equation*}$$ Let $xH\in G/H$. Then $x\in G$. Since $G=\langle a\rangle$, there exists $n\in\mathbb{Z}$ such that $$\begin{equation*} x=a^n. \end{equation*}$$ Therefore $$\begin{align*} xH &=a^nH\\ &=(aH)^n. \end{align*}$$ Thus every element of $G/H$ is a power of $aH$. Therefore $$\begin{equation*} G/H=\langle aH\rangle. \end{equation*}$$ Hence, $G/H$ is cyclic. $\square$ ::: :::corollary Let $(G,\circ)$ be a cyclic group and let $H$ be a subgroup of $G$. Then $G/H$ is abelian. ::: :::proof Given that $(G,\circ)$ is a cyclic group and $H$ is a subgroup of $G$. To prove that $G/H$ is abelian. Since $G$ is cyclic, the quotient group $G/H$ is cyclic. Since every cyclic group is abelian, $G/H$ is abelian. Hence, $G/H$ is abelian. $\square$ ::: The theorem does not say that every quotient of a non-cyclic group is non-cyclic. It only says that cyclicity of $G$ is sufficient for cyclicity of $G/H$. Quotients can simplify structure, so a group with complicated multiplication may still have a very simple quotient. This is one reason quotient groups are so useful in classification problems. We observe which properties are guaranteed to pass from $G$ to $G/H$. If $G$ is abelian, then every quotient is abelian. If $G$ is cyclic, then every quotient is cyclic, and therefore abelian. Turn the assumptions on and off to see that the theorems give sufficient conditions, not converses. :::scientific-preview[Quotient Property Explorer] ::: The arrows go from the original group to the quotient group. A quotient can gain simplicity, so an abelian quotient does not force the original group to be abelian. :::solved-problem Let $G=(\mathbb{Z},+)$ and let $H=n\mathbb{Z}$, where $n$ is a positive integer. Prove that $\mathbb{Z}/n\mathbb{Z}$ is cyclic. ::: :::solution Let $n$ be a positive integer. The group $(\mathbb{Z},+)$ is cyclic, generated by $1$. Since $\mathbb{Z}$ is abelian, $n\mathbb{Z}$ is normal in $\mathbb{Z}$. By the quotient of a cyclic group theorem, $$\begin{equation*} \mathbb{Z}/n\mathbb{Z} \end{equation*}$$ is cyclic. Indeed, $$\begin{equation*} \mathbb{Z}/n\mathbb{Z}=\langle 1+n\mathbb{Z}\rangle. \end{equation*}$$ ::: :::solved-problem Let $G=\langle a\rangle$ and let $H\leq G$. Show that $aH$ generates $G/H$. ::: :::solution Let $xH\in G/H$. Since $G=\langle a\rangle$, there exists $m\in\mathbb{Z}$ such that $$\begin{equation*} x=a^m. \end{equation*}$$ Therefore $$\begin{align*} xH &=a^mH\\ &=(aH)^m. \end{align*}$$ Thus every element of $G/H$ is a power of $aH$. Hence $$\begin{equation*} G/H=\langle aH\rangle. \end{equation*}$$ ::: We model quotients of cyclic groups using $\mathbb{Z}/n\mathbb{Z}$. Enter $n$ and a proposed generator coset $r+n\mathbb{Z}$. The calculator computes the order of that coset and decides whether it generates the whole quotient. This makes the theorem concrete: in $\mathbb{Z}/n\mathbb{Z}$, the coset $1+n\mathbb{Z}$ always generates. :::calculator[Cyclic Quotient Generator Calculator] ::: :::exercise 1. Prove that a quotient of an abelian group is abelian. 2. Prove that a quotient of a cyclic group is cyclic. 3. Find a generator of $\mathbb{Z}/5\mathbb{Z}$. 4. If $G=\langle a\rangle$ and $H\leq G$, identify a generator of $G/H$. 5. Does $G/H$ abelian imply that $G$ is abelian? ::: :::answer 1. If $G$ is abelian, then $(aH)(bH)=(ab)H=(ba)H=(bH)(aH)$. 2. If $G=\langle a\rangle$, then every coset $xH$ has the form $(aH)^n$, so $G/H=\langle aH\rangle$. 3. The coset $1+5\mathbb{Z}$ is a generator of $\mathbb{Z}/5\mathbb{Z}$. 4. The coset $aH$ generates $G/H$. 5. No. A quotient can be abelian even when the original group is not abelian. ::: :::faq[Frequently Asked Questions] Q: Is every quotient of an abelian group abelian? A: Yes. Every subgroup of an abelian group is normal, and coset multiplication remains commutative. Q: Is every quotient of a cyclic group cyclic? A: Yes. If $G=\langle a\rangle$, then $G/H=\langle aH\rangle$. Q: Can a non-abelian group have an abelian quotient? A: Yes. Quotients can collapse enough non-commutative behaviour to produce an abelian group. ::: :::call-to-action[Continue to Normal Subgroups in Quotient Groups] subtitle: Learn how normal subgroups correspond between a group and its quotient group. button: Next Lesson | published/abstract-algebra/subgroups-and-normal-subgroups/quotient-groups/normal-subgroups-in-quotient-groups.md button-ghost: Previous Lesson | published/abstract-algebra/subgroups-and-normal-subgroups/quotient-groups/construction-of-quotient-groups.md :::