# alternating group is a normal subgroup of the symmetric group

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\begin{theorem}
The alternating group $A_{n}$ is a normal subgroup of the symmetric group $S_{n}$
\end{theorem}

\begin{proof}
Define the epimorphism $f:S_{n}\rightarrow\mathbb{Z}_2$ by
$:\sigma\mapsto 0$ if $\sigma$ is an even permutation and
$:\sigma\mapsto 1$ if $\sigma$ is an odd permutation.  Hence,
$A_{n}$ is the kernel of $f$ and so it is a normal subgroup of the
domain $S_{n}$. Furthermore $S_{n}/A_{n}\cong\mathbb{Z}_2$ by
the first isomorphism theorem. So by Lagrange's theorem
$$\vert S_{n} \vert=\vert A_{n} \vert\vert S_{n}/A_{n}\vert.$$
Therefore, $\vert A_{n}\vert=n!/2$. That is, there are $n!/2$ many
elements in $A_{n}$
\end{proof}

\textbf{Remark}.  What we have shown in the theorem is that, in fact, $A_n$ has index $2$ in $S_n$.  In general, if a subgroup $H$ of $G$ has index $2$, then $H$ is normal in $G$. (Since $[G:H]=2$, there is an element $g\in G-H$, so that $gH\cap H=\varnothing$ and thus $gH=Hg$).
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