After proving the inverse product criterion, we now study a common sufficient condition for a group to be abelian. Suppose every element satisfies $a^2=e$. Then every element is its own inverse. This strong condition forces commutativity. Groups with this property appear naturally in examples such as the Klein four group and vector spaces over the field with two elements. In this lesson, students will prove that a group in which every element has square equal to the identity must be abelian. :::definition[Exponent Two Condition] Let $(G,\circ)$ be a group with identity element $e$. We say that $G$ satisfies the $exponent two condition$ if $$\begin{equation*} a^2=e \quad \forall\, a\in G. \end{equation*}$$ Equivalently, every element of $G$ is its own inverse. ::: If $a^2=e$, then $a\circ a=e$, so $a$ is an inverse of itself. By uniqueness of inverse, $a^{-1}=a$. Thus the exponent two condition says that inversion does not change any element. This is a strong restriction. It does not say that every element has order exactly $2$, because the identity has order $1$. It says that every non-identity element has order $2$. :::theorem Let $(G,\circ)$ be a group with identity element $e$. If $$\begin{equation*} a^2=e \quad \forall\, a\in G, \end{equation*}$$ then $G$ is abelian. ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$. Given that $$\begin{equation*} a^2=e \quad \forall\, a\in G. \end{equation*}$$ To prove that $G$ is abelian. Let $a,b\in G$. Since $a\circ b\in G$, the hypothesis gives $$\begin{equation*} (a\circ b)^2=e. \end{equation*}$$ Therefore $$\begin{align*} (a\circ b)^2=e &\implies a\circ b=(a\circ b)^{-1} \quad [\because\, a\circ b \text{ is its own inverse}]\\ &\implies a\circ b=b^{-1}\circ a^{-1} \quad [\because\, \text{inverse of a product}]\\ &\implies a\circ b=b\circ a \quad [\because\, a^{-1}=a,\; b^{-1}=b]. \end{align*}$$ Therefore $$\begin{equation*} a\circ b=b\circ a \quad \forall\, a,b\in G. \end{equation*}$$ Hence, $G$ is abelian. $\square$ ::: The proof uses the hypothesis twice: once for the product $a\circ b$, and once for the elements $a$ and $b$ separately. This is why the condition must hold for every element of the group. It is not enough that each of $a$ and $b$ has square equal to $e$; we also need $(a\circ b)^2=e$. :::example The Klein four group $$\begin{equation*} V_4=\{e,a,b,c\} \end{equation*}$$ satisfies $$\begin{equation*} e^2=e,\quad a^2=e,\quad b^2=e,\quad c^2=e. \end{equation*}$$ Therefore $V_4$ satisfies the exponent two condition. By the theorem, $V_4$ is abelian. ::: :::example The group $(\mathbb{Z}_2,+_2)$ satisfies the additive form of the exponent two condition: $$\begin{equation*} 2\overline{x}=\overline{0}\quad \forall\, \overline{x}\in\mathbb{Z}_2. \end{equation*}$$ Thus every element is its own additive inverse. The group is abelian. ::: :::solved-problem Let $(G,\circ)$ be a group with identity element $e$. Suppose that $$\begin{equation*} a^2=e \quad \forall\, a\in G. \end{equation*}$$ Prove that $$\begin{equation*} a^{-1}=a \quad \forall\, a\in G. \end{equation*}$$ ::: :::solution Let $(G,\circ)$ be a group with identity element $e$. Given that $$\begin{equation*} a^2=e \quad \forall\, a\in G. \end{equation*}$$ To prove that $$\begin{equation*} a^{-1}=a \quad \forall\, a\in G. \end{equation*}$$ Let $a\in G$. Then $$\begin{equation*} a^2=e. \end{equation*}$$ Thus $$\begin{equation*} a\circ a=e. \end{equation*}$$ Also $a\circ a=e$ gives both left and right inverse equations for $a$ with itself. Therefore $a$ is an inverse of $a$. Since the inverse of $a$ is unique, $$\begin{equation*} a^{-1}=a. \end{equation*}$$ Hence, $$\begin{equation*} a^{-1}=a \quad \forall\, a\in G. \end{equation*}$$ $\square$ ::: :::solved-problem Let $(G,\circ)$ be a group with exactly two elements. Prove that every element of $G$ satisfies $x^2=e$. ::: :::solution Let $(G,\circ)$ be a group with identity element $e$ and exactly two elements. To prove that every element of $G$ satisfies $x^2=e$. Write $$\begin{equation*} G=\{e,a\}, \end{equation*}$$ where $a\neq e$. For $x=e$, $$\begin{equation*} e^2=e. \end{equation*}$$ Now consider $a$. Since $G$ is closed under $\circ$, $$\begin{equation*} a^2\in G. \end{equation*}$$ If possible let $$\begin{equation*} a^2=a. \end{equation*}$$ Then $$\begin{align*} a^2=a &\implies a\circ a=a\circ e \quad [\because\, a\circ e=a]\\ &\implies a=e \quad [\because\, \text{left cancellation law}]. \end{align*}$$ A contradiction. Hence, $$\begin{equation*} a^2=e. \end{equation*}$$ Therefore every element of $G$ satisfies $x^2=e$. $\square$ ::: :::corollary Let $(G,\circ)$ be a group. If $|G|=2$, then $G$ is abelian. ::: :::proof Given that $(G,\circ)$ is a group and $|G|=2$. To prove that $G$ is abelian. By the previous solved result, every element $x\in G$ satisfies $$\begin{equation*} x^2=e. \end{equation*}$$ By the exponent two theorem, $G$ is abelian. Hence, every group of order $2$ is abelian. $\square$ ::: :::remark The exponent two theorem gives a fast way to prove commutativity in many small groups. However, it is only a sufficient condition. There are many abelian groups that do not satisfy $a^2=e$ for all elements. For example, $(\mathbb{Z}_3,+_3)$ is abelian, but $\overline{1}+\overline{1}\neq\overline{0}$. ::: We observe the exponent two condition in additive notation by testing whether $2x=0$ for every element. In $\mathbb{Z}_n$ under addition, this means checking whether $2x\equiv 0\pmod n$ for all residues $x$. Use the value of $n$ to test the whole group, not only one element. Try setting $n=2$, $n=4$, and $n=6$ and observe which elements fail the condition. :::calculator[Exponent Two Test] ::: Notice that $\mathbb{Z}_2$ satisfies the condition because every residue is its own additive inverse. For larger cyclic groups, at least one residue usually fails $2x=0$, showing that abelian groups need not satisfy the exponent two condition. :::exercise [1] State the exponent two condition. [2] Prove that if $a^2=e$ for every $a\in G$, then $a^{-1}=a$ for every $a\in G$. [3] Prove that a group satisfying the exponent two condition is abelian. [4] Give an example of an abelian group that does not satisfy the exponent two condition. [5] Prove that every group of order $2$ is abelian. ::: :::answer [1] The condition is $a^2=e$ for every $a\in G$. [2] Since $a\circ a=e$, the element $a$ is its own inverse. [3] For $a,b\in G$, $(a\circ b)^2=e$, so $a\circ b=(a\circ b)^{-1}=b^{-1}\circ a^{-1}=b\circ a$. [4] The group $(\mathbb{Z}_3,+_3)$ is abelian but does not satisfy $2x=0$ for every element. [5] If $G=\{e,a\}$, then $a^2=e$, so the group satisfies the exponent two condition and is abelian. ::: :::faq Q: What does $a^2=e$ for every element mean? A: It means every element is its own inverse. Q: Does the exponent two condition imply abelian? A: Yes. Every group satisfying it is abelian. Q: Does every abelian group satisfy the exponent two condition? A: No. Many abelian groups have elements of order greater than $2$. Q: Does the identity have order $2$? A: No. The identity has order $1$. ::: :::call-to-action[Power Criteria for Commutativity] subtitle: Continue to power identities that force a group to be abelian. button: Next Lesson | published/abstract-algebra/introduction-to-groups/abelian-groups/power-criteria-for-abelian-groups button-ghost: Previous Lesson | published/abstract-algebra/introduction-to-groups/abelian-groups/inverse-product-criterion :::