After learning the symmetric group, we now study one of its most important subgroups: the alternating group. The alternating group consists of even permutations. It is a standard finite group that appears throughout abstract algebra, especially in permutation theory, group actions, and the study of solvability. The smallest nontrivial alternating group, $\mathbb{A}_3$, has only three elements, while $\mathbb{A}_n$ has $\dfrac{n!}{2}$ elements for $n\geq2$. In this lesson, students will verify that alternating groups are finite groups under composition. :::definition[Even Permutation] Let $\sigma\in\mathbb{S}_n$. The permutation $\sigma$ is called an $even permutation$ if it can be expressed as a product of an even number of transpositions. ::: :::definition[Alternating Group] Let $n\in\mathbb{N}$. The $alternating group$ $\mathbb{A}_n$ is the set of all even permutations in $\mathbb{S}_n$. Thus $$\begin{equation*} \mathbb{A}_n=\{\sigma\in\mathbb{S}_n:\sigma \text{ is even}\}. \end{equation*}$$ The operation on $\mathbb{A}_n$ is composition of permutations. ::: The parity of a permutation is well-defined: a permutation cannot be both even and odd. This fact allows us to separate $\mathbb{S}_n$ into even and odd permutations. For $n\geq2$, exactly half of the permutations are even and half are odd, so $$\begin{equation*} |\mathbb{A}_n|=\frac{n!}{2}. \end{equation*}$$ The identity permutation is even because it is a product of zero transpositions. :::theorem Let $n\geq2$. Then $(\mathbb{A}_n,\circ)$ is a finite group under composition of permutations, and $$\begin{equation*} |\mathbb{A}_n|=\frac{n!}{2}. \end{equation*}$$ ::: :::proof Given that $n\geq2$ and $\mathbb{A}_n$ is the set of all even permutations in $\mathbb{S}_n$. To prove that $(\mathbb{A}_n,\circ)$ is a finite group. [1] To prove closure. Let $\sigma,\tau\in\mathbb{A}_n$. Then $\sigma$ and $\tau$ are even permutations. A product of two even permutations is even because the total number of transpositions in the combined expression is even. Therefore $$\begin{equation*} \sigma\circ\tau\in\mathbb{A}_n. \end{equation*}$$ [2] To prove associativity. Let $\sigma,\tau,\rho\in\mathbb{A}_n$. Since composition of functions is associative, $$\begin{equation*} (\sigma\circ\tau)\circ\rho=\sigma\circ(\tau\circ\rho). \end{equation*}$$ [3] To prove the existence of identity element. The identity permutation $I$ is even because it is a product of zero transpositions. Hence $I\in\mathbb{A}_n$. Also $$\begin{equation*} \sigma\circ I=I\circ\sigma=\sigma \quad \forall\, \sigma\in\mathbb{A}_n. \end{equation*}$$ Therefore $I$ is the identity element. [4] To prove the existence of inverse elements. Let $\sigma\in\mathbb{A}_n$. Then $\sigma$ is even. If $\sigma$ is expressed as a product of an even number of transpositions, then $\sigma^{-1}$ is obtained by reversing the order of those transpositions. The number of transpositions remains even. Therefore $\sigma^{-1}$ is even and $$\begin{equation*} \sigma^{-1}\in\mathbb{A}_n. \end{equation*}$$ Also $$\begin{equation*} \sigma\circ\sigma^{-1}=\sigma^{-1}\circ\sigma=I. \end{equation*}$$ Thus every element has an inverse in $\mathbb{A}_n$. Since half of the $n!$ permutations in $\mathbb{S}_n$ are even for $n\geq2$, we get $$\begin{equation*} |\mathbb{A}_n|=\frac{n!}{2}. \end{equation*}$$ Hence, $(\mathbb{A}_n,\circ)$ is a finite group. $\square$ ::: :::example The group $\mathbb{A}_3$ is $$\begin{equation*} \mathbb{A}_3=\{I,(123),(132)\}. \end{equation*}$$ It has $$\begin{equation*} |\mathbb{A}_3|=\frac{3!}{2}=3 \end{equation*}$$ elements. The two $3$-cycles are even because each $3$-cycle can be written as a product of two transpositions. ::: :::example The group $\mathbb{A}_4$ has $$\begin{equation*} |\mathbb{A}_4|=\frac{4!}{2}=12 \end{equation*}$$ elements. It contains the identity permutation, all $3$-cycles in $\mathbb{S}_4$, and the products of two disjoint transpositions such as $(12)(34)$. ::: :::solved-problem Prove that $\mathbb{A}_3$ is a finite group under composition. ::: :::solution Given that $$\begin{equation*} \mathbb{A}_3=\{I,(123),(132)\}. \end{equation*}$$ To prove that $(\mathbb{A}_3,\circ)$ is a finite group. Since $\mathbb{A}_3$ consists of the even permutations in $\mathbb{S}_3$, closure follows because the composition of two even permutations is even. Associativity follows from associativity of function composition. The identity permutation $I$ belongs to $\mathbb{A}_3$ and satisfies $$\begin{equation*} \sigma\circ I=I\circ\sigma=\sigma \quad \forall\, \sigma\in\mathbb{A}_3. \end{equation*}$$ The inverse of $I$ is $I$, the inverse of $(123)$ is $(132)$, and the inverse of $(132)$ is $(123)$. Thus every element has an inverse in $\mathbb{A}_3$. Since $\mathbb{A}_3$ has $3$ elements, it is finite. Hence, $(\mathbb{A}_3,\circ)$ is a finite group. $\square$ ::: :::solved-problem Find the order of $(123)$ in $\mathbb{A}_3$. ::: :::solution Let $\sigma=(123)\in\mathbb{A}_3$. To find the order of $\sigma$. We compute $$\begin{equation*} \sigma^1=(123), \end{equation*}$$ $$\begin{equation*} \sigma^2=(132), \end{equation*}$$ and $$\begin{equation*} \sigma^3=I. \end{equation*}$$ Since no smaller positive power of $\sigma$ is $I$, the least positive integer $n$ such that $\sigma^n=I$ is $3$. Hence, $$\begin{equation*} o((123))=3. \end{equation*}$$ $\square$ ::: :::remark The alternating group $\mathbb{A}_n$ is not merely a subset of $\mathbb{S}_n$; it is a group in its own right under the same operation. The main verification points are closure under composition and closure under inverses. Both rely on parity of permutations. If a proof simply says that $\mathbb{A}_n$ is inside $\mathbb{S}_n$, it is incomplete because a subset of a group need not be a group. ::: The alternating group contains exactly the even permutations inside $\mathbb{S}_n$. We observe the total number $n!$ of permutations and the half-size $n!/2$ of $\mathbb{A}_n$. Change $n$ and notice how quickly the size grows. This reinforces that $\mathbb{A}_n$ is finite and that it is a large subgroup of $\mathbb{S}_n$ for bigger $n$. :::scientific-preview[Alternating Group Size Explorer] @libraries: plotly ::: Use the calculator to compute the size of $\mathbb{A}_n$ and test the parity of a permutation of $\{1,2,3\}$. Enter $n$ for the size formula and an image list such as $2,3,1$ for the parity check. The parity is computed by counting inversions in the image list. :::calculator[Alternating Group Size] ::: :::exercise [1] Define an even permutation. [2] Define $\mathbb{A}_n$. [3] Find $|\mathbb{A}_5|$. [4] List the elements of $\mathbb{A}_3$. [5] Find the inverse of $(123)$ in $\mathbb{A}_3$. ::: :::answer [1] An even permutation is a permutation expressible as a product of an even number of transpositions. [2] $\mathbb{A}_n$ is the set of all even permutations in $\mathbb{S}_n$. [3] $|\mathbb{A}_5|=5!/2=60$. [4] $\mathbb{A}_3=\{I,(123),(132)\}$. [5] The inverse of $(123)$ is $(132)$. ::: :::faq Q: What is $\mathbb{A}_n$? A: It is the group of all even permutations in $\mathbb{S}_n$. Q: What is the operation in $\mathbb{A}_n$? A: The operation is composition of permutations. Q: How many elements are in $\mathbb{A}_n$? A: For $n\geq2$, $|\mathbb{A}_n|=n!/2$. Q: Is $\mathbb{A}_3$ a group? A: Yes. It has three elements: $I,(123),(132)$. ::: :::call-to-action[Klein Four Group] subtitle: Continue to the Klein four group, a small but important abelian group of order four. button: Next Lesson | /abstract-algebra/introduction-to-groups/standard-examples-of-groups/klein-four-group button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/standard-examples-of-groups/permutation-groups :::