After studying alternating groups, we now finish this section with one of the most useful small examples in group theory: the Klein four group. This group has four elements, every non-identity element has order $2$, and the group is abelian. It is small enough to understand completely from its operation table, but important enough to appear repeatedly in finite group theory, symmetry, and direct product constructions. In this lesson, students will verify the group axioms for the Klein four group and compare it with cyclic groups of order four. :::definition[Klein Four Group] The $Klein four group$ is the set $$\begin{equation*} V_4=\{e,a,b,c\} \end{equation*}$$ with operation $\circ$ defined by the table $$\begin{equation*} \begin{array}{c|cccc} \circ & e & a & b & c\\ \hline e & e & a & b & c\\ a & a & e & c & b\\ b & b & c & e & a\\ c & c & b & a & e \end{array} \end{equation*}$$ ::: The table gives the entire operation. The first row and first column show that $e$ behaves as an identity element. The entries $a\circ a=e$, $b\circ b=e$, and $c\circ c=e$ show that each non-identity element is its own inverse. The table is symmetric across the main diagonal, so the operation is commutative. The only group axiom that is not visually immediate from the table is associativity; for this example, associativity may be verified directly from the table or recognized from a standard model of $V_4$. :::theorem The structure $(V_4,\circ)$ defined by the table above is a finite abelian group. ::: :::proof Given that $$\begin{equation*} V_4=\{e,a,b,c\} \end{equation*}$$ with the operation defined by the stated table. To prove that $(V_4,\circ)$ is a finite abelian group. [1] To prove closure. Every entry in the operation table belongs to $V_4$. Therefore $V_4$ is closed under $\circ$. [2] To prove associativity. We use the standard model $$\begin{equation*} V_4=\mathbb{Z}_2\times\mathbb{Z}_2 \end{equation*}$$ with componentwise addition modulo $2$, under the correspondence $$\begin{equation*} e=(\overline{0},\overline{0}),\quad a=(\overline{1},\overline{0}),\quad b=(\overline{0},\overline{1}),\quad c=(\overline{1},\overline{1}). \end{equation*}$$ The table is exactly the componentwise addition table under this correspondence. Since addition modulo $2$ is associative in each coordinate, componentwise addition is associative. Therefore $\circ$ is associative on $V_4$. [3] To prove the existence of identity element. From the table, $$\begin{equation*} e\circ x=x\circ e=x \quad \forall\, x\in V_4. \end{equation*}$$ Therefore $e$ is the identity element. [4] To prove the existence of inverse elements. From the table, $$\begin{equation*} e\circ e=e,\quad a\circ a=e,\quad b\circ b=e,\quad c\circ c=e. \end{equation*}$$ Thus every element is its own inverse. [5] To prove commutativity. The table is symmetric, so $$\begin{equation*} x\circ y=y\circ x \quad \forall\, x,y\in V_4. \end{equation*}$$ Therefore $V_4$ is abelian. Since $V_4$ has four elements, it is finite. Hence, $(V_4,\circ)$ is a finite abelian group. $\square$ ::: The proof of associativity is often the delicate part. Checking all triples directly would require $4^3=64$ computations. The model $\mathbb{Z}_2\times\mathbb{Z}_2$ gives a cleaner verification. Under componentwise addition modulo $2$, associativity is inherited from addition modulo $2$. This also explains why every non-identity element has order $2$. :::example In $V_4$, the order of $a$ is $2$ because $$\begin{equation*} a^2=a\circ a=e \end{equation*}$$ and $a\neq e$. Similarly, $$\begin{equation*} o(b)=2,\quad o(c)=2. \end{equation*}$$ The identity element has order $1$. ::: :::example The Klein four group is not cyclic. If $V_4$ were cyclic, it would have an element of order $4$. But the orders of the elements are $$\begin{equation*} o(e)=1,\quad o(a)=2,\quad o(b)=2,\quad o(c)=2. \end{equation*}$$ No element has order $4$. Therefore $V_4$ is not cyclic. ::: :::solved-problem Using the table of $V_4$, compute $a\circ b$, $b\circ c$, and $c\circ a$. ::: :::solution Given the Klein four group table. To compute $a\circ b$, $b\circ c$, and $c\circ a$. From the row of $a$ and column of $b$, we get $$\begin{equation*} a\circ b=c. \end{equation*}$$ From the row of $b$ and column of $c$, we get $$\begin{equation*} b\circ c=a. \end{equation*}$$ From the row of $c$ and column of $a$, we get $$\begin{equation*} c\circ a=b. \end{equation*}$$ Hence, $$\begin{equation*} a\circ b=c,\quad b\circ c=a,\quad c\circ a=b. \end{equation*}$$ $\square$ ::: :::solved-problem Prove that every non-identity element of $V_4$ has order $2$. ::: :::solution Let $$\begin{equation*} V_4=\{e,a,b,c\}. \end{equation*}$$ To prove that every non-identity element of $V_4$ has order $2$. The non-identity elements are $a,b,c$. From the operation table, $$\begin{equation*} a\circ a=e,\quad b\circ b=e,\quad c\circ c=e. \end{equation*}$$ Also, $$\begin{equation*} a\neq e,\quad b\neq e,\quad c\neq e. \end{equation*}$$ Therefore each of $a,b,c$ has least positive power $2$ equal to the identity. Thus $$\begin{equation*} o(a)=o(b)=o(c)=2. \end{equation*}$$ Hence, every non-identity element of $V_4$ has order $2$. $\square$ ::: :::remark There are two common groups of order $4$: the cyclic group of order $4$ and the Klein four group. They are not the same type of group. A cyclic group of order $4$ has an element of order $4$. The Klein four group has no element of order $4$. This distinction will become important when group isomorphism is introduced. ::: The Klein four group can be modeled as $\mathbb{Z}_2\times\mathbb{Z}_2$ with componentwise addition. We observe the four elements as points in a two-bit square and compute products by adding coordinates modulo $2$. Choose two elements and notice that the result is the fourth corner when the two non-identity elements are different. This visualization explains both commutativity and the fact that every non-identity element has order $2$. :::scientific-preview[Klein Four Operation Explorer] @libraries: plotly ::: Use the calculator to multiply elements in the Klein four group. Choose two elements and read the product from the table. The calculator also reports the order of the product, so you can compare the identity element with the three non-identity elements. :::calculator[Klein Four Product] ::: :::exercise [1] Write the elements of $V_4$. [2] What is the identity element of $V_4$? [3] Find the inverse of $a$ in $V_4$. [4] Find $b\circ c$ in $V_4$. [5] Prove that $V_4$ is not cyclic. ::: :::answer [1] $V_4=\{e,a,b,c\}$. [2] The identity element is $e$. [3] The inverse of $a$ is $a$. [4] $b\circ c=a$. [5] The orders are $1,2,2,2$, so no element has order $4$. Hence $V_4$ is not cyclic. ::: :::faq Q: How many elements does the Klein four group have? A: It has four elements. Q: Is the Klein four group abelian? A: Yes. Its operation table is symmetric. Q: What is the order of each non-identity element? A: Each non-identity element has order $2$. Q: Is the Klein four group cyclic? A: No. It has no element of order $4$. ::: :::call-to-action[Cyclic Groups] subtitle: Continue to cyclic groups, where generators and powers describe the entire group. button: Next Section | /abstract-algebra/introduction-to-groups/cyclic-groups/cyclic-group button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/standard-examples-of-groups/alternating-groups :::