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For an exercise, I need to prove that the alternating group $A_n$ is a normal subgroup of the symmetric group $S_n$.

As clue they say we can use a group homomorphism $\operatorname{sgn} : S_n \to \{-1,1\}$. I really don't see how i can use this.... can somebody help?

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    $\begingroup$ Does $\ker(sgn) = A_n$ help you? $\endgroup$ – Moritz Mar 9 '15 at 19:06
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    $\begingroup$ yeah, it would but why is ker(sgn)=$A_n$? $\endgroup$ – Robbe Motmans Mar 9 '15 at 19:10
  • $\begingroup$ Because every element of $A_n$ is mapped to $1$ and every other element in $S_n\setminus A_n$ to $-1$. Note that $1$ is group-unity of $(\{\pm 1\},\cdot)$. $\endgroup$ – Moritz Mar 9 '15 at 19:12
  • $\begingroup$ oke thnx alot guys, i get it know i think. $\endgroup$ – Robbe Motmans Mar 9 '15 at 19:13
  • $\begingroup$ Final hint: $\ker(\psi)$ for any group-homomorphism $\psi: G \to H$ between groups $G$ and $H$ is always a normal subgroup of $G$. $\endgroup$ – Moritz Mar 9 '15 at 19:14
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$1$.Note that kernal of sign homomorphism is precisely $A_n$ and kernal of a homomorphism is a normal subgroup.

$2$. Recall that every Subgroup of index 2 is Normal and note that $[S_n:A_n]=2$

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  • $\begingroup$ oke thnx, then i can prove it! $\endgroup$ – Robbe Motmans Mar 9 '15 at 19:08
  • $\begingroup$ do you also know how i can use the given group homomorphism to prove the same? $\endgroup$ – Robbe Motmans Mar 9 '15 at 19:09
  • $\begingroup$ Hint:kernal of given map is $A_n$ $\endgroup$ – Arpit Kansal Mar 9 '15 at 19:10
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Add another method to prove in addition to Arpit's answer: conjugation preserves cycle type; so $s a s^{-1} \in A_n$

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