# alternating group has index 2 in the symmetric group, the

We prove that the alternating group^{} ${A}_{n}$ has index 2 in the symmetric group^{} ${S}_{n}$, i.e., ${A}_{n}$ has the same cardinality as its complement ${S}_{n}\setminus {A}_{n}$. The proof is function-theoretic. Its idea is similar to the proof in the parent topic, but the focus is less on algebraic aspect.

Let $\pi \in {S}_{n}\setminus {A}_{n}$. Define $\pi :{S}_{n}\setminus {A}_{n}\to {A}_{n}$ by $\pi (\sigma )=\pi \sigma $, where $\pi \sigma $ is the product^{} of $\pi $ and $\sigma $.

$$\pi (\sigma )=\pi (\delta )\u27f9\sigma =\delta $$ |

since ${\pi}^{-1}$ exists and ${\pi}^{-1}\pi \sigma ={\pi}^{-1}\pi \delta $.

Onto: Given $\alpha \in {A}_{n}$, there exists an element in ${S}_{n}\setminus {A}_{n}$, namely $\lambda ={\pi}^{-1}\alpha $, such that

$$\pi (\alpha )=\lambda .$$ |

(The element $\lambda $ is in ${S}_{n}\setminus {A}_{n}$ because ${\pi}^{-1}$ is and the product of an odd permutation^{} and an even permutation is odd.)

The function $\pi :{S}_{n}\setminus {A}_{n}\to {A}_{n}$ is, therefore, a one-to-one correspondence, so both sets ${S}_{n}\setminus {A}_{n}$ and ${A}_{n}$ have the same cardinality.

Title | alternating group has index 2 in the symmetric group, the |
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Canonical name | AlternatingGroupHasIndex2InTheSymmetricGroupThe |

Date of creation | 2013-03-22 16:48:49 |

Last modified on | 2013-03-22 16:48:49 |

Owner | yesitis (13730) |

Last modified by | yesitis (13730) |

Numerical id | 8 |

Author | yesitis (13730) |

Entry type | Proof |

Classification | msc 20-00 |