[proofplan]
We use the signature map to identify $A_n$ as the kernel of a homomorphism. Normality is then proved directly by conjugating an element of $A_n$ and using multiplicativity of the signature. For the order computation, we fix one transposition and use left multiplication by it to construct a bijection from even permutations to odd permutations, so the two classes have equal cardinality.
[/proofplan]
[step:Identify $A_n$ as the kernel of the signature homomorphism]
By [citetheorem:7877], the map
\begin{align*}
\operatorname{sgn}:S_n\to\{1,-1\}
\end{align*}
is a [group homomorphism](/page/Group%20Homomorphism), where $\{1,-1\}$ is equipped with multiplication. By the definition of $A_n$,
\begin{align*}
A_n=\{\sigma\in S_n:\operatorname{sgn}(\sigma)=1\}=\ker(\operatorname{sgn}).
\end{align*}
Thus $A_n$ is the kernel of the [signature homomorphism](/theorems/7877).
[/step]
[step:Prove normality by conjugating inside the kernel]
Let $\rho\in S_n$ and let $\alpha\in A_n$. Since $\alpha\in A_n$, we have $\operatorname{sgn}(\alpha)=1$. Since $\operatorname{sgn}$ is a homomorphism by [citetheorem:7877],
\begin{align*}
\operatorname{sgn}(\rho\alpha\rho^{-1})=\operatorname{sgn}(\rho)\operatorname{sgn}(\alpha)\operatorname{sgn}(\rho^{-1}).
\end{align*}
Also,
\begin{align*}
1=\operatorname{sgn}(\operatorname{id}_{\{1,\dots,n\}})=\operatorname{sgn}(\rho\rho^{-1})=\operatorname{sgn}(\rho)\operatorname{sgn}(\rho^{-1}).
\end{align*}
Substituting $\operatorname{sgn}(\alpha)=1$ gives
\begin{align*}
\operatorname{sgn}(\rho\alpha\rho^{-1})=1.
\end{align*}
Therefore $\rho\alpha\rho^{-1}\in A_n$. Since $\rho\in S_n$ and $\alpha\in A_n$ were arbitrary, $A_n\trianglelefteq S_n$.
[guided]
We want to prove that $A_n$ is normal in $S_n$. By definition, this means that for every $\rho\in S_n$ and every $\alpha\in A_n$, the conjugate $\rho\alpha\rho^{-1}$ again lies in $A_n$.
The condition for membership in $A_n$ is expressed through the signature: an element $\beta\in S_n$ lies in $A_n$ exactly when $\operatorname{sgn}(\beta)=1$. Thus we compute the signature of the conjugate. Since [citetheorem:7877] says that
\begin{align*}
\operatorname{sgn}:S_n\to\{1,-1\}
\end{align*}
is a homomorphism, we may multiply signatures across products:
\begin{align*}
\operatorname{sgn}(\rho\alpha\rho^{-1})=\operatorname{sgn}(\rho)\operatorname{sgn}(\alpha)\operatorname{sgn}(\rho^{-1}).
\end{align*}
Because $\alpha\in A_n$, its signature is $1$. It remains to handle the two outer factors. Applying the homomorphism property to $\rho\rho^{-1}=\operatorname{id}_{\{1,\dots,n\}}$ gives
\begin{align*}
1=\operatorname{sgn}(\operatorname{id}_{\{1,\dots,n\}})=\operatorname{sgn}(\rho\rho^{-1})=\operatorname{sgn}(\rho)\operatorname{sgn}(\rho^{-1}).
\end{align*}
Therefore
\begin{align*}
\operatorname{sgn}(\rho\alpha\rho^{-1})=\operatorname{sgn}(\rho)\cdot 1\cdot \operatorname{sgn}(\rho^{-1})=1.
\end{align*}
So $\rho\alpha\rho^{-1}\in A_n$. Since this holds for every $\rho\in S_n$ and every $\alpha\in A_n$, the subgroup $A_n$ is normal in $S_n$.
[/guided]
[/step]
[step:Pair even permutations with odd permutations using a fixed transposition]
Since $n\ge 2$, the transposition $\tau=(1\ 2)$ belongs to $S_n$. By [citetheorem:7875],
\begin{align*}
\operatorname{sgn}(\tau)=-1.
\end{align*}
Define the set of odd permutations
\begin{align*}
O_n:=\{\sigma\in S_n:\operatorname{sgn}(\sigma)=-1\}.
\end{align*}
Define the map
\begin{align*}
L_{\tau}:A_n\to O_n
\end{align*}
by $L_{\tau}(\sigma)=\tau\sigma$ for each $\sigma\in A_n$.
The map is well-defined: if $\sigma\in A_n$, then by [citetheorem:7877],
\begin{align*}
\operatorname{sgn}(L_{\tau}(\sigma))=\operatorname{sgn}(\tau\sigma)=\operatorname{sgn}(\tau)\operatorname{sgn}(\sigma)=(-1)\cdot 1=-1.
\end{align*}
Hence $L_{\tau}(\sigma)\in O_n$.
Define another map
\begin{align*}
M_{\tau}:O_n\to A_n
\end{align*}
by $M_{\tau}(\omega)=\tau\omega$ for each $\omega\in O_n$. If $\omega\in O_n$, then
\begin{align*}
\operatorname{sgn}(M_{\tau}(\omega))=\operatorname{sgn}(\tau\omega)=\operatorname{sgn}(\tau)\operatorname{sgn}(\omega)=(-1)(-1)=1,
\end{align*}
so $M_{\tau}(\omega)\in A_n$. Since $\tau^2=\operatorname{id}_{\{1,\dots,n\}}$, for every $\sigma\in A_n$,
\begin{align*}
M_{\tau}(L_{\tau}(\sigma))=\tau(\tau\sigma)=\sigma,
\end{align*}
and for every $\omega\in O_n$,
\begin{align*}
L_{\tau}(M_{\tau}(\omega))=\tau(\tau\omega)=\omega.
\end{align*}
Thus $L_{\tau}$ is a bijection from $A_n$ to $O_n$, with inverse $M_{\tau}$.
[/step]
[step:Count the two parity classes and conclude the order formula]
Every permutation in $S_n$ has signature either $1$ or $-1$, so
\begin{align*}
S_n=A_n\cup O_n
\end{align*}
and the union is disjoint. The previous step gives $|A_n|=|O_n|$. Since $|S_n|=n!$, finite additivity of cardinality for disjoint finite sets gives
\begin{align*}
n!=|S_n|=|A_n|+|O_n|=2|A_n|.
\end{align*}
Dividing by $2$ yields
\begin{align*}
|A_n|=\frac{n!}{2}.
\end{align*}
Together with the normality proved above, this proves $A_n\trianglelefteq S_n$ and $|A_n|=n!/2$.
[/step]
