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How to prove that $A_n$ is normal in $S_n$?

Note that $A_n$ is a group of even permutations on a set of length $n$. $S_n$ is the group of all permutations on $n$ symbols.

Definition: A subgroup $H$ is normal in $G$ means: for all $g \in G$, $g^{-1}Hg\subseteq H$.

Andrés E. Caicedo
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Melissa Herrington
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    It is (by definition) the kernel of the sign homomorphism. – MooS Dec 02 '15 at 08:21
  • @MooS An answer (a good and rigourous one moreover) should be posted as an answer. – Pece Dec 02 '15 at 08:22
  • You can show this in various way: The solution of @MooS is elegant, you can also try to calculate the cardinality of $A_n$ and see that is a half than the cardinality of $S_n$. You can try also to notice that the conjugacy classes are of the same order and are formed by k-cycle, and so every cojugation remains inside the k-cycles for every k – Alessio Ranallo Dec 02 '15 at 08:26

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Subgroup of index $2$ is Normal.

R.N
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    As sneaky as this proof is, it somewhat detracts from the group structure of $A_n$ in $S_n$. Indeed, one can show a bijective correspondence between $A_n$ and $S_n \setminus A_n$ (multiplication by a transposition will do) and thereby invoke the group structure. – user217285 Dec 02 '15 at 08:50
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Hint:

Based on the definitions you have provided you should try to show that if $g$ is any permutation (odd or even) then $g^{-1}$ will be the same. Now if $h$ is a permutation, then the parity of the number of transpositions required to represent $ghg^{-1}$ will be the same as that of $h$.

Anurag A
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Let $\epsilon: (S_n, \circ) \to (\{-1,1\}, \times)$ be the signature map; i.e.:

For a given $p \in S_n$, $\epsilon(p) = (-1)^{\large \# I_p}$, where $I_p:=\{(i,j), i < j \text{ and } p(i) > p(j) \}$.

We have that $\epsilon$ is a homomorphism; meaning that for any two permutations $p_1$ and $p_2$, $\epsilon(p_1 p_2) = \epsilon(p_1) \epsilon(p_2)$.

Observe that $\ker \epsilon = \{ p \in S_n, \epsilon(p) = 1 \} = A_n$. As the kernel of a homomorphism is a normal subgroup, it follows that $A_n \unlhd S_n$.