The Alternating Groups, An

Recall that if we have the finite $n$-element set $\{1, 2, ..., n \}$ then $S_n$ is the set of all permutations of elements from $\{1, 2, ..., n \}$ known as the symmetric group on $n$ elements as we saw earlier on The Symmetric Groups, Sn page.

On the Even and Odd Cycles page, we noted that every permutation $\sigma \in S_n$ is either defined to be even (if $\sigma$ can be decomposed into a product of an even number of transpositions) or odd (if $\sigma$ can be decomposed into a product of an odd number of transpositions). We even saw on The Identity Permutation page that the identity permutation $\epsilon$ is an even permutation.

Now consider the subset $A_n \subseteq S_n$ which we define as:

(1)

\begin{align} \quad A_n = \{ \sigma \in S_n : \sigma \: \mathrm{is \: an \: even \: permutation \: of \: the \: elements \: in} \: \{1, 2, ..., n \} \} \end{align}

We will see that the set $A_n$ forms a group with respect to the operation $\circ$ which we first define below.

Definition: If $A_n = \{ \sigma \in S_n : \sigma \: \mathrm{is \: an \: even \: permutation \: of \: the \: elements \: in} \: \{1, 2, ..., n \} \}$ then $(A_n, \circ)$ is the Alternating Group on $n$-elements.

We now prove that $(A_n, \circ)$ is indeed a group.

Proposition 1: The set $A_n$ of even permutations of elements in $\{ 1, 2, ..., n \}$ forms a group with respect to the operation $\circ$ of function composition.

• Proof: Since $A_n \subseteq S_n$, we only need to show that $A_n$ is closed under $\circ$ and that for each element $\sigma \in A_n$ there exists $\sigma^{-1} \in A_n$ such that $\sigma \circ \sigma^{-1} = \epsilon$ and $\sigma^{-1} \circ \sigma = \epsilon$ where $\epsilon$ is the identity permutation.

• Let $\sigma, \gamma \in A_n$. Then $\sigma = t_1 \circ t_2 \circ ... \circ t_k$ and $\gamma = u_1 \circ u_2 \circ ... \circ u_j$ where $t_1, t_2, ..., t_k, u_1, u_2, ..., u_j$ are transpositions and such that $k$ and $j$ are even. Then:

(2)

\begin{align} \quad \sigma \circ \gamma = (t_1 \circ t_2 \circ ... \circ t_k) \circ (u_1 \circ u_2 \circ ... \circ u_j) \end{align}

• Then $\sigma \circ \gamma$ can be written as the product of $k + j$ transpositions. Since $k$ and $j$ are both even numbers, we have that their sum $k + j$ is even. Therefore $(\sigma \circ \gamma) \in A_n$.

• Now let $\sigma \in A_n$. Then $\sigma$ can be decomposed into an even number of transpositions, say $\sigma = t_1 \circ t_2 \circ ... \circ t_k$ where $k$ is even.. Since $\sigma$ is a permutation, it is bijective, and so there exists $\sigma^{-1} \in S_n$ such that $\sigma \circ \sigma^{-1} = \epsilon$ and $\sigma^{-1} \circ \sigma = \epsilon$. We must then only show that $\sigma^{-1}$ is contained in $A_n$. Suppose that $\sigma^{-1} \not \in A_n$. Then $\sigma^{-1}$ can be decomposed into an odd number of transpositions, say $\sigma^{-1} = u_1 \circ u_2 \circ ... \circ u_j$ where $j$ is odd. Then:

(3)

\begin{align} \quad \epsilon = \sigma \circ \sigma^{-1} = (t_1 \circ t_2 \circ ... \circ t_k) \circ (u_1 \circ u_2 \circ ... \circ u_j) \end{align}

• Hence $\epsilon$ can be decomposed into a product of $k + j$ transpositions. Since $k$ is even and $j$ is odd, we have that $k + j$ is odd. But we've proven that the identity permutation $\epsilon$ is an even permutation, so we've arrived at a contradiction. Hence the assumption that $\sigma^{-1} \not \in A_n$ was false. So for all $\sigma \in A_n$ we have that $\sigma^{-1} \in A_n$ is such that $\sigma \circ \sigma^{-1} = \epsilon$ and $\sigma^{-1} \circ \sigma = \epsilon$.

• Therefore $(A_n, \circ)$ is a subgroup of $(S_n, \circ)$ and hence $(A_n, \circ)$ is a group itself. $\blacksquare$