After studying groups in which every element has square equal to the identity, we now look at power identities that force commutativity. These tests are useful because some identities look like ordinary exponent rules, but in a non-abelian group they are not automatically valid. For example, $(a\circ b)^2=a^2\circ b^2$ is not true in every group. In this lesson, students will prove that certain power identities imply that the group is abelian, and they will see how cancellation is used to recover $a\circ b=b\circ a$. :::theorem Let $(G,\circ)$ be a group. If $$\begin{equation*} (a\circ b)^2=a^2\circ b^2 \quad \forall\, a,b\in G, \end{equation*}$$ then $G$ is abelian. ::: :::proof Given that $(G,\circ)$ is a group. Given that $$\begin{equation*} (a\circ b)^2=a^2\circ b^2 \quad \forall\, a,b\in G. \end{equation*}$$ To prove that $G$ is abelian. Let $a,b\in G$. Then $$\begin{align*} (a\circ b)^2=a^2\circ b^2 &\implies a\circ b\circ a\circ b=a\circ a\circ b\circ b\\ &\implies b\circ a\circ b=a\circ b\circ b \quad [\because\, \text{left cancellation law}]\\ &\implies b\circ a=a\circ b \quad [\because\, \text{right cancellation law}]. \end{align*}$$ Therefore $$\begin{equation*} a\circ b=b\circ a \quad \forall\, a,b\in G. \end{equation*}$$ Hence, $G$ is abelian. $\square$ ::: The theorem says that the familiar rule $(ab)^2=a^2b^2$ is a commutativity test. In an abelian group it holds because $ab=ba$. Conversely, if it holds for every pair, then the middle factors are forced to commute. This is a common place where students import a rule from number algebra without checking whether the group operation is commutative. :::corollary Let $(G,\circ)$ be a group. Then $G$ is abelian if and only if $$\begin{equation*} (a\circ b)^2=a^2\circ b^2 \quad \forall\, a,b\in G. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group. To prove that $G$ is abelian if and only if $$\begin{equation*} (a\circ b)^2=a^2\circ b^2 \quad \forall\, a,b\in G. \end{equation*}$$ [1] If $G$ is abelian, then for all $a,b\in G$, $$\begin{align*} (a\circ b)^2 &=a\circ b\circ a\circ b\\ &=a\circ a\circ b\circ b \quad [\because\, a\circ b=b\circ a]\\ &=a^2\circ b^2. \end{align*}$$ [2] If $(a\circ b)^2=a^2\circ b^2$ for all $a,b\in G$, then the previous theorem gives that $G$ is abelian. Hence, the equivalence holds. $\square$ ::: :::theorem Let $(G,\circ)$ be a group. If $$\begin{equation*} (a\circ b)^3=a^3\circ b^3 \quad \forall\, a,b\in G \end{equation*}$$ and $$\begin{equation*} (a\circ b)^5=a^5\circ b^5 \quad \forall\, a,b\in G, \end{equation*}$$ then $G$ is abelian. ::: :::proof Given that $(G,\circ)$ is a group. Given that $$\begin{equation*} (a\circ b)^3=a^3\circ b^3 \quad \forall\, a,b\in G \end{equation*}$$ and $$\begin{equation*} (a\circ b)^5=a^5\circ b^5 \quad \forall\, a,b\in G. \end{equation*}$$ To prove that $G$ is abelian. Let $a,b\in G$. From the cube identity, $$\begin{align*} (a\circ b)^3=a^3\circ b^3 &\implies a\circ b\circ a\circ b\circ a\circ b=a\circ a\circ a\circ b\circ b\circ b\\ &\implies b\circ a\circ b\circ a=a^2\circ b^2 \quad [\because\, \text{left and right cancellation laws}]. \end{align*}$$ From the fifth-power identity, $$\begin{align*} (a\circ b)^5=a^5\circ b^5 &\implies b\circ a\circ b\circ a\circ b\circ a\circ b\circ a=a^4\circ b^4 \quad [\because\, \text{left and right cancellation laws}]. \end{align*}$$ Using $$\begin{equation*} b\circ a\circ b\circ a=a^2\circ b^2 \end{equation*}$$ inside the previous line, we get $$\begin{align*} (a^2\circ b^2)\circ(a^2\circ b^2)&=a^4\circ b^4\\ a^2\circ b^2\circ a^2\circ b^2&=a^2\circ a^2\circ b^2\circ b^2\\ b^2\circ a^2\circ b^2&=a^2\circ b^2\circ b^2 \quad [\because\, \text{left cancellation law}]\\ b^2\circ a^2&=a^2\circ b^2 \quad [\because\, \text{right cancellation law}]. \end{align*}$$ Therefore $$\begin{equation*} b\circ a\circ b\circ a=b^2\circ a^2. \end{equation*}$$ Thus $$\begin{align*} b\circ a\circ b\circ a=b\circ b\circ a\circ a &\implies a\circ b\circ a=b\circ a\circ a \quad [\because\, \text{left cancellation law}]\\ &\implies a\circ b=b\circ a \quad [\because\, \text{right cancellation law}]. \end{align*}$$ Therefore $$\begin{equation*} a\circ b=b\circ a \quad \forall\, a,b\in G. \end{equation*}$$ Hence, $G$ is abelian. $\square$ ::: The proof corrects a common gap in this type of problem. One cannot simply state that two power identities imply commutativity. The proof must reduce the identities through cancellation until the expression $a\circ b=b\circ a$ appears. The cube identity gives a four-factor relation, and the fifth-power identity turns that relation into a square-type commutativity relation for $a^2$ and $b^2$. A final cancellation then gives the desired commutativity of $a$ and $b$. :::solved-problem Let $(G,\circ)$ be a group. If $$\begin{equation*} (a\circ b)^2=a^2\circ b^2 \quad \forall\, a,b\in G, \end{equation*}$$ prove that $$\begin{equation*} (a\circ b)^n=a^n\circ b^n \end{equation*}$$ for every $n\in\mathbb{N}$ and all $a,b\in G$. ::: :::solution Let $(G,\circ)$ be a group. Given that $$\begin{equation*} (a\circ b)^2=a^2\circ b^2 \quad \forall\, a,b\in G. \end{equation*}$$ To prove that $$\begin{equation*} (a\circ b)^n=a^n\circ b^n \end{equation*}$$ for every $n\in\mathbb{N}$. By the square criterion, $G$ is abelian. Since $G$ is abelian, powers of products distribute over factors. Thus for $n\in\mathbb{N}$, $$\begin{align*} (a\circ b)^n &=\underbrace{(a\circ b)\circ(a\circ b)\circ\cdots\circ(a\circ b)}_{n\text{ times}}\\ &=a^n\circ b^n. \end{align*}$$ Hence, $$\begin{equation*} (a\circ b)^n=a^n\circ b^n. \end{equation*}$$ $\square$ ::: :::remark A power identity may be true for a particular pair of elements without making the group abelian. The theorem requires the identity to hold for all $a,b\in G$. The word “all” is essential because abelian is a global property of the group operation. ::: We observe the square identity by expanding $(a\circ b)^2$ as $a\circ b\circ a\circ b$. The obstruction is the middle pair $b\circ a$, because changing $a\circ b\circ a\circ b$ into $a\circ a\circ b\circ b$ requires $b\circ a=a\circ b$. Use the selector to compare the non-abelian expansion, the abelian rearrangement, and the cancellation argument. This makes the square identity a test for commutativity rather than an automatic exponent rule. :::calculator[Square Power Test] ::: Notice that the square identity works because cancellation exposes the middle factor swap. Once $b\circ a=a\circ b$ is forced for arbitrary $a$ and $b$, the group is abelian by definition. :::exercise [1] Prove that if $(a\circ b)^2=a^2\circ b^2$ for all $a,b\in G$, then $G$ is abelian. [2] If $G$ is abelian, prove that $(a\circ b)^3=a^3\circ b^3$. [3] State the cube-and-fifth-power criterion from this lesson. [4] Explain why the identity $(a\circ b)^2=a^2\circ b^2$ is not automatic in a non-abelian group. [5] If $(a\circ b)^2=a^2\circ b^2$ for all $a,b\in G$, simplify $(a\circ b)^7$. ::: :::answer [1] Expanding and cancelling gives $a\circ b\circ a\circ b=a\circ a\circ b\circ b$, hence $b\circ a=a\circ b$. [2] Since $a\circ b=b\circ a$, we can rearrange $a\circ b\circ a\circ b\circ a\circ b$ as $a^3\circ b^3$. [3] If $(a\circ b)^3=a^3\circ b^3$ and $(a\circ b)^5=a^5\circ b^5$ for all $a,b\in G$, then $G$ is abelian. [4] Because $(a\circ b)^2=a\circ b\circ a\circ b$, and changing it to $a\circ a\circ b\circ b$ requires $a\circ b=b\circ a$. [5] Since the condition makes $G$ abelian, $(a\circ b)^7=a^7\circ b^7$. ::: :::faq Q: Is $(a\circ b)^2=a^2\circ b^2$ always true? A: No. It is true for all pairs exactly when the group is abelian. Q: Why does the square identity force commutativity? A: Expanding both sides and using cancellation gives $b\circ a=a\circ b$. Q: Do the cube and fifth power identities together force commutativity? A: Yes, when both identities hold for every pair of group elements. Q: Are these criteria finite-group results? A: No. They hold for arbitrary groups. ::: :::call-to-action[Elements of Order Two] subtitle: Continue to involutions, products of commuting elements of order two, and centrality of a unique element of order two. button: Next Lesson | published/abstract-algebra/introduction-to-groups/abelian-groups/elements-of-order-two button-ghost: Previous Lesson | published/abstract-algebra/introduction-to-groups/abelian-groups/groups-of-exponent-two :::