This is exercise 9.6b from Section. The Rational Numbers in textbook Analysis I by Amann/Escher.
On the symmetric group $\mathrm{S}_n$, define the sign function by $$\operatorname{sign} \sigma := \prod_{1 \leq j<k \leq n} \frac{\sigma(j)-\sigma(k)}{j-k}, \qquad \sigma \in \mathrm{S}_{n}$$
The alternating group $\mathrm{A}_{n}$ is defined as $\mathrm{A}_{n} := \{\sigma \in \mathrm{S}_{n} \mid \operatorname{sign} \sigma = 1\}$.
A transposition is a permutation which interchanges two numbers and leaves the others fixed.
Prove that
$\mathrm{A}_{n}$ has order $n!/2$ for $n \geq 2$, and $1$ for $n=1$.
For $n \ge 2$, any permutation $\sigma \in \mathrm{S}_{n}$ can be represented as a composition of transpositions: $\sigma=\sigma_{1} \circ \sigma_{2} \circ \cdots \circ \sigma_{N}$, and then $\operatorname{sign} \sigma=(-1)^{N}$, independent of this representation. Thus the number of transpositions in the representation is even for even permutations and odd for odd permutations.
Could you please verify if my attempt is fine or contains logical gaps/errors?
My attempt:
- For $\sigma \in \mathrm{A}_{n}$, we define $\sigma' \in \mathrm{S}_{n}$ by $\sigma' (k) = \sigma (k)$ for all $k \in \{2 , \ldots,n\}$, $\sigma' (1) = \sigma (2)$, and $\sigma' (2) = \sigma (1)$. It follows that $\operatorname{sign} \sigma' = -1$ and that the function
$$\begin {array}{lrcl} & \mathrm{A}_{n} & \longrightarrow & \mathrm{S}_{n} - \mathrm{A}_{n} \\ & \sigma & \longmapsto & \sigma' \end{array}$$
is bijective. Then $|\mathrm{A}_{n}| = |\mathrm{S}_{n} - \mathrm{A}_{n}| = |\mathrm{S}_{n}|/2 = n!/2$ for $n \ge 2$.
- For $\sigma \in \mathrm{S}_{n}$, we define $(\sigma_m)_{m \le n}$ recursively by $\sigma_0 = \operatorname{id}_{\{1,\ldots,n\}}$ and $\sigma_{m}$ by transposing $m$ and ${(\sigma_0 \circ \cdots \circ \sigma_{m-1})}^{-1} \circ \sigma (m)$.
Next we prove $\sigma_0 \circ \cdots \circ \sigma_{m} \restriction \{1, \ldots, m\} = \sigma \restriction \{1, \ldots, m\}$ by induction on $m$. Since $\sigma_1$ transposes $1$ and $\sigma_0^{-1} \circ \sigma(1) = \sigma(1)$. It follows that the statement holds for $m = 1$. Let it hold for $m$. Then $\sigma_0 \circ \cdots \circ \sigma_{m} \restriction \{1, \ldots, m\} = \sigma \restriction \{1, \ldots, m\}$. Because $\sigma_{m+1}$ transposes ${m+1}$ and ${(\sigma_0 \circ \cdots \circ \sigma_{m})}^{-1} \circ \sigma ({m+1})$, we have $$\begin{aligned}\sigma_0 \circ \cdots \circ \sigma_{m} \circ \sigma_{m+1} (m+1) &= \sigma_0 \circ \cdots \circ \sigma_{m} \left ({(\sigma_0 \circ \cdots \circ \sigma_{m})}^{-1} \circ \sigma ({m+1}) \right)\\ &= {(\sigma_0 \circ \cdots \circ \sigma_{m})} \circ {(\sigma_0 \circ \cdots \circ \sigma_{m})}^{-1} \circ \sigma ({m+1})\\ &= \sigma ({m+1})\end{aligned}$$
I claim that ${(\sigma_0 \circ \cdots \circ \sigma_{m})}^{-1} \circ \sigma ({m+1}) > m$. If not, ${(\sigma_0 \circ \cdots \circ \sigma_{m})}^{-1} \circ \sigma ({m+1}) = m' \le m$. Then $\sigma ({m+1}) = \sigma_0 \circ \cdots \circ \sigma_{m} (m') = \sigma (m')$ by inductive hypothesis. This contradicts the fact that $\sigma$ is bijective.
If $i \le m$ then $m+1 \neq i < {(\sigma_0 \circ \cdots \circ \sigma_{m})}^{-1} \circ \sigma ({m+1})$ and thus $\sigma_0 \circ \cdots \circ \sigma_{m} \circ \sigma_{m+1} (i) = \sigma_0 \circ \cdots \circ \sigma_{m} (i) = \sigma (i)$ by inductive hypothesis.
As a result, $$\sigma_0 \circ \cdots \circ \sigma_{m+1} \restriction \{1, \ldots, m+1\} = \sigma \restriction \{1, \ldots, m+1\}$$
This completes the proof. It follows directly that $$\begin{aligned}\sigma_1 \circ \cdots \circ \sigma_{n} &= \sigma_0 \circ \cdots \circ \sigma_{n} \restriction \{1, \ldots, n\} \\ &= \sigma \restriction \{1, \ldots, n\} \\ &= \sigma \end{aligned}$$
where $\sigma_1, \ldots, \sigma_n$ are transpositions.