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This is not too obvious to me - what is the size of alternating group?

Following the hint in the comment, should it be $A_n = S_n/2$?

So I don't feel right up to here.....

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    $\begingroup$ Use the fact that $|S_n| = n!$ and that $[S_n:A_n] = 2$ $\endgroup$ – Prahlad Vaidyanathan Sep 30 '13 at 15:27
  • $\begingroup$ Oh... So following that, we get $A_n = n!/2 ?$ Thank you so much @PrahladVaidyanathan $\endgroup$ – Tumbleweed Sep 30 '13 at 15:30
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The map $\sigma:S_n\to\mathbb{Z}/2\mathbb{Z}$ defined by sending a permutation to $0$ if it has even parity, and $1$ if it has odd parity, is a group homomorphism. The kernel of this map is $A_n$, so by the first isomorphism theorem, we have $[S_n:A_n]=2$ for $n\ge 2$ (the map is not surjective for $n=1$). It follows that for $n\ge 2$ we have

$$|A_n|=\frac{|S_n|}{2}=\frac{n!}{2}$$

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The key is to use the existence of the sign homomorphism $\text{sgn} : S_n \rightarrow \{ \pm 1 \}$. By definition $A_n$ is the kernel of $\text{sgn}$. Since $\text{sgn}$ is surjective, it follows immediately that $[S_n : A_n ] = 2$, so $|A_n| = |S_n| / 2 = n! / 2$.

Edit: As noted by Jared, one must assume that $n \geq 2$.

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