After studying power criteria for commutativity, we finish the abelian-groups section with elements of order two. Such elements are often called involutions. They are important because they are their own inverses and frequently control commutativity patterns in small groups. When two distinct elements of order two commute, their product also has order two. When an element of order two is unique in a group, it must commute with every element of the group. In this lesson, students will prove these order-two results carefully. :::definition[Element of Order Two] Let $(G,\circ)$ be a group with identity element $e$. An element $a\in G$ is called an $element of order two$ if $$\begin{equation*} o(a)=2. \end{equation*}$$ Equivalently, $$\begin{equation*} a\neq e \end{equation*}$$ and $$\begin{equation*} a^2=e. \end{equation*}$$ ::: An element of order two is its own inverse. Indeed, if $a^2=e$, then $a\circ a=e$, so $a^{-1}=a$. In permutation groups, transpositions are elements of order two. In the Klein four group, every non-identity element has order two. In a cyclic group of even order, there is exactly one element of order two. :::theorem Let $(G,\circ)$ be a group with identity element $e$, and let $a,b\in G$. If $a\neq b$, $o(a)=o(b)=2$, and $$\begin{equation*} a\circ b=b\circ a, \end{equation*}$$ then $$\begin{equation*} o(a\circ b)=2. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$, $a,b\in G$, $a\neq b$, $o(a)=o(b)=2$, and $a\circ b=b\circ a$. To prove that $$\begin{equation*} o(a\circ b)=2. \end{equation*}$$ Since $o(a)=2$ and $o(b)=2$, we get $$\begin{equation*} a^2=e,\quad b^2=e. \end{equation*}$$ Now $$\begin{align*} (a\circ b)^2 &=a\circ b\circ a\circ b\\ &=a\circ a\circ b\circ b \quad [\because\, a\circ b=b\circ a]\\ &=a^2\circ b^2\\ &=e\circ e\\ &=e. \end{align*}$$ Also $a\circ b\neq e$. If possible let $$\begin{equation*} a\circ b=e. \end{equation*}$$ Then $$\begin{align*} a\circ b=e &\implies a=b^{-1}\\ &\implies a=b \quad [\because\, o(b)=2]. \end{align*}$$ A contradiction, since $a\neq b$. Therefore $a\circ b\neq e$. Since $(a\circ b)^2=e$ and $a\circ b\neq e$, we get $$\begin{equation*} o(a\circ b)=2. \end{equation*}$$ Hence, $o(a\circ b)=2$. $\square$ ::: The commuting hypothesis is necessary in the first theorem. Without $a\circ b=b\circ a$, we cannot rewrite $a\circ b\circ a\circ b$ as $a\circ a\circ b\circ b$. In non-abelian groups, products of order-two elements can have order greater than two. This is one of the first places where non-commutativity changes the behaviour of order. :::theorem Let $(G,\circ)$ be a group with identity element $e$, and let $a,b\in G$. If $o(b)=2$, then $$\begin{equation*} o(a\circ b\circ a^{-1})=2. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$, $a,b\in G$, and $o(b)=2$. To prove that $$\begin{equation*} o(a\circ b\circ a^{-1})=2. \end{equation*}$$ Since $o(b)=2$, we get $$\begin{equation*} b^2=e \end{equation*}$$ and $b\neq e$. Now $$\begin{align*} (a\circ b\circ a^{-1})^2 &=(a\circ b\circ a^{-1})\circ(a\circ b\circ a^{-1})\\ &=a\circ b\circ(a^{-1}\circ a)\circ b\circ a^{-1}\\ &=a\circ b\circ e\circ b\circ a^{-1}\\ &=a\circ b^2\circ a^{-1}\\ &=a\circ e\circ a^{-1}\\ &=e. \end{align*}$$ Also $a\circ b\circ a^{-1}\neq e$. If possible let $$\begin{equation*} a\circ b\circ a^{-1}=e. \end{equation*}$$ Multiplying on the left by $a^{-1}$ and on the right by $a$, we get $$\begin{align*} a\circ b\circ a^{-1}=e &\implies b=a^{-1}\circ e\circ a\\ &\implies b=e, \end{align*}$$ which contradicts $o(b)=2$. Therefore $a\circ b\circ a^{-1}\neq e$. Hence, $$\begin{equation*} o(a\circ b\circ a^{-1})=2. \end{equation*}$$ $\square$ ::: :::corollary Let $(G,\circ)$ be a group with identity element $e$, and let $a\in G$. If $a$ is the only element of order two in $G$, then $$\begin{equation*} a\circ x=x\circ a \quad \forall\, x\in G. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$, $a\in G$, and $a$ is the only element of order two in $G$. To prove that $$\begin{equation*} a\circ x=x\circ a \quad \forall\, x\in G. \end{equation*}$$ Let $x\in G$. Since $o(a)=2$, the previous theorem gives $$\begin{equation*} o(x\circ a\circ x^{-1})=2. \end{equation*}$$ Since $a$ is the only element of order two, we get $$\begin{equation*} x\circ a\circ x^{-1}=a. \end{equation*}$$ Multiplying on the right by $x$, we obtain $$\begin{align*} x\circ a\circ x^{-1}=a &\implies x\circ a=a\circ x. \end{align*}$$ Therefore $$\begin{equation*} a\circ x=x\circ a \quad \forall\, x\in G. \end{equation*}$$ Hence, $a$ commutes with every element of $G$. $\square$ ::: The corollary says that a unique element of order two is central. In other words, it commutes with every group element. This result does not say that the whole group is abelian. It says that this particular element must lie in the commutative part of the group. Later, this idea is described using the center of a group. :::example In the cyclic group $\langle a\rangle$ of order $8$, the unique element of order two is $a^4$. Therefore $a^4$ commutes with every element of the group. Since the group is cyclic, all elements commute anyway, but the uniqueness result gives a more general reason for this one element. ::: :::example In the Klein four group $V_4$, there are three elements of order two. Therefore no single non-identity element is the unique element of order two. The uniqueness corollary does not apply. The group is still abelian, but for a different reason: its operation table is symmetric. ::: :::solved-problem Let $(G,\circ)$ be a group and let $a,b\in G$ with $a\neq b$, $o(a)=o(b)=2$, and $a\circ b=b\circ a$. Prove that $a\circ b$ is not equal to $a$ or $b$. ::: :::solution Let $(G,\circ)$ be a group and let $a,b\in G$ with $a\neq b$, $o(a)=o(b)=2$, and $a\circ b=b\circ a$. To prove that $a\circ b\neq a$ and $a\circ b\neq b$. If possible let $$\begin{equation*} a\circ b=a. \end{equation*}$$ Then $$\begin{align*} a\circ b=a &\implies a\circ b=a\circ e \quad [\because\, a\circ e=a]\\ &\implies b=e \quad [\because\, \text{left cancellation law}], \end{align*}$$ which contradicts $o(b)=2$. Therefore $a\circ b\neq a$. If possible let $$\begin{equation*} a\circ b=b. \end{equation*}$$ Then $$\begin{align*} a\circ b=b &\implies a\circ b=e\circ b \quad [\because\, e\circ b=b]\\ &\implies a=e \quad [\because\, \text{right cancellation law}], \end{align*}$$ which contradicts $o(a)=2$. Hence, $a\circ b\neq a$ and $a\circ b\neq b$. $\square$ ::: We observe products of order-two elements inside the Klein four group $V_4=\{e,a,b,c\}$. The non-identity elements $a$, $b$, and $c$ all have order two, and the group is abelian. Select two elements and multiply them. Try choosing two distinct elements to see a product of order two, then choose the same element twice to see why the distinctness condition matters. :::calculator[Order Two Product] ::: Notice that distinct commuting involutions in $V_4$ multiply to another involution. Choosing the same involution twice gives the identity instead, so the product has order one rather than order two. :::exercise [1] Define an element of order two. [2] Prove that an element of order two is its own inverse. [3] If $a\neq b$, $o(a)=o(b)=2$, and $a\circ b=b\circ a$, prove that $o(a\circ b)=2$. [4] If $o(b)=2$, prove that $o(a\circ b\circ a^{-1})=2$. [5] If a group has a unique element of order two, prove that this element commutes with every group element. ::: :::answer [1] It is a non-identity element $a$ such that $a^2=e$. [2] Since $a\circ a=e$, the element $a$ is its own inverse. [3] Compute $(a\circ b)^2=a\circ a\circ b\circ b=e$, and show $a\circ b\neq e$. [4] $(a\circ b\circ a^{-1})^2=a\circ b^2\circ a^{-1}=e$, and the conjugate is not $e$. [5] For any $x\in G$, the conjugate $x\circ a\circ x^{-1}$ also has order two; uniqueness gives $x\circ a\circ x^{-1}=a$, hence $x\circ a=a\circ x$. ::: :::faq Q: What is an element of order two? A: It is a non-identity element $a$ such that $a^2=e$. Q: Is an element of order two its own inverse? A: Yes. If $a^2=e$, then $a^{-1}=a$. Q: Does the product of two order-two elements always have order two? A: Not always. The result in this lesson assumes the two elements commute and are distinct. Q: What happens if there is exactly one element of order two? A: That element commutes with every element of the group. ::: :::call-to-action[Subgroups] subtitle: Continue to subgroups after the uploaded abelian-group material has been completed. button: Next Section | published/abstract-algebra/introduction-to-groups/subgroups/subgroup-definition button-ghost: Previous Lesson | published/abstract-algebra/introduction-to-groups/abelian-groups/power-criteria-for-abelian-groups :::