Proving $|A_n|=n!/2$, where $A_n$ is the alternating group.

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In the book: Theory and Problems Abstract Algebra Schaum's Outline Series by J. FANG:

He proved that $|A_n|=n!/2$, but in proof there is one thing that I did not understand, here is proof.(page: 83)

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He first found $s\le r$ but how? He defined all odd permutations are $q_1,q_2,...,q_s$ and when you multiply even permutations, with a transposition you get odd permutations (we cannot say: all the odd permutations) , so I think it must $r\le s$ not $s\le r$. Moreover, the same logic follows in the last paragraph $r\le s$, again, it must $s\le r$.

What do you think whether FANG, or me is right?