After counting quotient groups, we now study when a quotient group is abelian. The quotient $G/H$ is abelian when the failure of elements of $G$ to commute is absorbed by $H$. This failure is measured by commutators. Thus abelian quotient groups do not require $G$ itself to be abelian; they require all commutators to fall inside the normal subgroup being factored out. This lesson is a first step toward the commutator subgroup and abelianisation, although we keep the discussion at the undergraduate quotient-group level. :::definition[Commutator] Let $(G,\circ)$ be a group and let $x,y\in G$. The commutator of $x$ and $y$ is the element $$\begin{equation*} x\circ y\circ x^{-1}\circ y^{-1}. \end{equation*}$$ ::: The commutator equals the identity element when $x$ and $y$ commute. Indeed, $x\circ y=y\circ x$ is equivalent to $x\circ y\circ x^{-1}\circ y^{-1}=e$. Therefore a commutator records exactly how far a pair of elements is from commuting. In a quotient group, the identity element is the coset $H$, so a commutator lying in $H$ becomes invisible in $G/H$. :::theorem Let $(G,\circ)$ be a group and let $H$ be a normal subgroup of $G$. Then $G/H$ is abelian if and only if $$\begin{equation*} x\circ y\circ x^{-1}\circ y^{-1}\in H \quad \forall x,y\in G. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group and $H$ is a normal subgroup of $G$. To prove that $G/H$ is abelian if and only if $x\circ y\circ x^{-1}\circ y^{-1}\in H$ for all $x,y\in G$. [1] Let $G/H$ be abelian. Let $x,y\in G$. Since $G/H$ is abelian, $$\begin{equation*} (xH)(yH)=(yH)(xH). \end{equation*}$$ Therefore $$\begin{equation*} (x\circ y)H=(y\circ x)H. \end{equation*}$$ Thus $$\begin{equation*} (y\circ x)^{-1}\circ(x\circ y)\in H. \end{equation*}$$ This gives $$\begin{equation*} x^{-1}\circ y^{-1}\circ x\circ y\in H. \end{equation*}$$ Since $H$ is normal in $G$, conjugating this element by $x\circ y$ gives another element of $H$. Now $$\begin{align*} (x\circ y)\circ(x^{-1}\circ y^{-1}\circ x\circ y)\circ(x\circ y)^{-1} &=x\circ y\circ x^{-1}\circ y^{-1}. \end{align*}$$ Therefore $$\begin{equation*} x\circ y\circ x^{-1}\circ y^{-1}\in H. \end{equation*}$$ [2] Let $$\begin{equation*} x\circ y\circ x^{-1}\circ y^{-1}\in H \quad \forall x,y\in G. \end{equation*}$$ To prove that $G/H$ is abelian. Let $xH,yH\in G/H$. By the given condition, $$\begin{equation*} x\circ y\circ x^{-1}\circ y^{-1}\in H. \end{equation*}$$ Since $H$ is normal in $G$, conjugating by $(y\circ x)^{-1}$ gives $$\begin{equation*} (y\circ x)^{-1}\circ(x\circ y)\in H. \end{equation*}$$ Therefore $$\begin{equation*} (x\circ y)H=(y\circ x)H. \end{equation*}$$ Thus $$\begin{align*} (xH)(yH) &=(x\circ y)H\\ &=(y\circ x)H\\ &=(yH)(xH). \end{align*}$$ Therefore $G/H$ is abelian. Hence, $G/H$ is abelian if and only if $x\circ y\circ x^{-1}\circ y^{-1}\in H$ for all $x,y\in G$. $\square$ ::: :::lemma Let $(G,\circ)$ be a group. If $x^2=e$ for every $x\in G$, then $G$ is abelian. ::: :::proof Given that $(G,\circ)$ is a group and $x^2=e$ for every $x\in G$. To prove that $G$ is abelian. Let $a,b\in G$. Since $a^2=e$ and $b^2=e$, we get $$\begin{equation*} a^{-1}=a \end{equation*}$$ and $$\begin{equation*} b^{-1}=b. \end{equation*}$$ Since $(a\circ b)^2=e$, we get $$\begin{equation*} (a\circ b)^{-1}=a\circ b. \end{equation*}$$ But $$\begin{equation*} (a\circ b)^{-1}=b^{-1}\circ a^{-1}=b\circ a. \end{equation*}$$ Therefore $$\begin{equation*} a\circ b=b\circ a. \end{equation*}$$ Since $a,b\in G$ were arbitrary, $G$ is abelian. Hence, $G$ is abelian. $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $$\begin{equation*} x^2\in H \quad \forall x\in G, \end{equation*}$$ then $G/H$ is abelian. ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $x^2\in H$ for every $x\in G$. To prove that $G/H$ is abelian. Since $x^2\in H$ for every $x\in G$, the subgroup $H$ is normal in $G$. Therefore $G/H$ is a quotient group. Let $xH\in G/H$. Then $$\begin{align*} (xH)^2 &=x^2H\\ &=H. \end{align*}$$ Thus every element of $G/H$ has square equal to the identity element $H$. By the exponent two lemma, $G/H$ is abelian. Hence, $G/H$ is abelian. $\square$ ::: We observe how commutators decide whether a quotient is abelian. Mark whether all commutators of $G$ lie in $H$ and whether every square lies in $H$. The first condition is exactly the abelian quotient criterion. The square condition is a useful sufficient condition because it makes every element of $G/H$ have order at most two. :::scientific-preview[Commutator Quotient Explorer] ::: Inside $G/H$, every element of $H$ is identified with the identity coset. When all commutators land in $H$, every failure of commutativity disappears in the quotient. :::solved-problem Let $H\trianglelefteq G$. Prove that if every commutator $x\circ y\circ x^{-1}\circ y^{-1}$ lies in $H$, then $G/H$ is abelian. ::: :::solution Let $xH,yH\in G/H$. Since $x\circ y\circ x^{-1}\circ y^{-1}\in H$ for all $x,y\in G$, the commutator criterion gives $$\begin{equation*} G/H \text{ is abelian}. \end{equation*}$$ Equivalently, $$\begin{align*} (xH)(yH) &=(x\circ y)H\\ &=(y\circ x)H\\ &=(yH)(xH). \end{align*}$$ ::: We use checklist-style input because the theorem is about universal conditions. Select the hypotheses you know. The calculator reports whether the quotient is proved abelian by the commutator criterion or by the square condition. It does not claim the quotient is non-abelian when the sufficient data is missing. :::calculator[Abelian Quotient Criterion Calculator] ::: :::exercise 1. Define the commutator of $x$ and $y$ in a group $G$. 2. Prove that $x$ and $y$ commute if and only if their commutator is $e$. 3. State the commutator criterion for $G/H$ to be abelian. 4. Prove that if $x^2\in H$ for every $x\in G$, then $G/H$ is abelian. 5. Give a reason why $G/H$ can be abelian even when $G$ is not abelian. ::: :::answer 1. The commutator is $x\circ y\circ x^{-1}\circ y^{-1}$. 2. If $x\circ y=y\circ x$, then $x\circ y\circ x^{-1}\circ y^{-1}=e$. Conversely, if the commutator is $e$, then $x\circ y=y\circ x$. 3. If $H\trianglelefteq G$, then $G/H$ is abelian if and only if every commutator $x\circ y\circ x^{-1}\circ y^{-1}$ lies in $H$. 4. The condition gives $H\trianglelefteq G$, and every element of $G/H$ has square $H$. A group in which every element has square equal to the identity is abelian. 5. The non-commuting part of $G$ may lie inside $H$, so it becomes the identity in $G/H$. ::: :::faq[Frequently Asked Questions] Q: What does a commutator measure? A: It measures the failure of two elements to commute. Q: Why does a commutator in $H$ become harmless in $G/H$? A: Elements of $H$ belong to the identity coset of the quotient group. Q: Does $G/H$ abelian imply that $G$ is abelian? A: No. It only says that all commutators of $G$ lie inside $H$. ::: :::call-to-action[Continue to Quotients by Central Subgroups] subtitle: Study central quotient groups and the theorem that cyclic central quotients force abelian groups. button: Next Lesson | published/abstract-algebra/subgroups-and-normal-subgroups/quotient-groups/quotients-by-central-subgroups.md button-ghost: Previous Lesson | published/abstract-algebra/subgroups-and-normal-subgroups/quotient-groups/order-of-a-quotient-group.md :::