Timeline for An easy proof that $S(n)$ does not embed into $A(n+1)$?

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May 16, 2011 at 16:53	comment	added	Arturo Magidin		@Derek: For that matter, cosets are not defined until page 24, right after this set of exercises, and Lagrange appears in page 26. The immediately previous exercise consists of showing that the 3-cycles generate $A_n$ for $n\gt 2$, so perhaps the argument Rotman has in mind consists of showing somehow that any putative image of $S_n$ in $A_{n+1}$ would necessarily contain all $3$-cycles.
May 16, 2011 at 16:31	comment	added	Arturo Magidin		@Gerry: Actually, that is the immediately previous exercise to the one in question. (The one Len is asking about is 2.8; proving that $A_n$ is generated by 3-cycles for $n\gt 2$ is Exercise 2.7).
May 16, 2011 at 12:29	comment	added	Gerry Myerson		That $A_{n+1}$ is generated by the 3-cycles - has that been established by page 22?
May 16, 2011 at 10:34	comment	added	Derek Holt		Yes but the whole point was to avoid using the simplicity of $A_{n+1}$, which has certainly not been covered by Page 22 of Rotman's book.
May 16, 2011 at 10:27	comment	added	Johannes Hahn		You don't need Betran's Postulat for that argument: Small n can be done by hand and for $n\geq 4$ the group $A_{n+1}$ is simple, so that the homomorphism $A_{n+1}\to S_m$ must be injective. Now $m! < \frac{(n+1)!}{2}$ and that's the contradiction.
May 16, 2011 at 10:10	history	answered	Derek Holt	CC BY-SA 3.0