The Size of the Alternating Groups

Recall from the The Alternating Groups, An page that if $A_n = \{ \sigma \in S_n : \sigma \: \mathrm{is \: an \: even \: permutation \: of \:} \{ 1, 2, ..., n \} \}$ then $(A_n, \circ)$ is called the alternating group on $n$ elements.

We know that $(A_n, \circ)$ is a subgroup of the symmetric group on $n$ elements, $(S_n, \circ)$ and that $\mid S_n \mid = n!$. So, we would now like to determine $\mid A_n \mid$. Since $A_n$ is the set of all even permutations on $\{ 1, 2, ..., n \}$ and every permutation is either even or odd, we would expect that $\displaystyle{\mid A_n \mid = \frac{n!}{2}}$. Fortunately this is indeed the case as we prove below.

Proposition 1: The size/order of $(A_n, \circ)$ is $\displaystyle{\mid A_n \mid = \frac{n!}{2}}$.

• Proof: Let $B_n \subset S_n$ be the set of all odd permutations of $\{ 1, 2, ..., n \}$ and let $\tau$ be any transposition. Define a function $f : A_n \to B_n$ for all $\sigma \in A_n$ by

(1)

\begin{align} \quad f(\sigma) = \tau \circ \sigma \end{align}

• Note that $\tau \circ \sigma$ is an odd permutation. If we can show that $f$ is a bijection then we must have that $\mid A_n \mid = \mid B_n \mid$. We first show that $f$ is injective. Let $\sigma_1, \sigma_2 \in A_n$ and suppose that $f(\sigma_1) = f(\sigma_2)$. Then:

(2)

\begin{align} \quad \tau \sigma_1 = \tau \sigma_2 \end{align}

• By the cancellation property we have that then $\sigma_1 = \sigma_2$ so $f$ is injective.

• Now we show that $f$ is surjective. Let $\sigma \in B_n$. Then $\tau \sigma$ is an even permutation. Noting that $\tau \circ \tau = \epsilon$ we have that:

(3)

\begin{align} \quad f (\tau \circ \sigma) = \tau \circ \tau \circ \sigma = \epsilon \circ \sigma = \sigma \end{align}

• So for every $\sigma \in B_n$ there exists a $\tau \circ \sigma \in A_n$ such that $f(\tau \circ \sigma) = \sigma$ so $f$ is surjective.

• Hence $\mid A_n \mid = \mid B_n \mid$. So $2 \mid A_n \mid = n!$ and $\displaystyle{\mid A_n \mid = \frac{n!}{2}}$. $\blacksquare$