-2

For the alternating group $A_4$, why is there no subgroup of order $6$? Also, what is a good way to explain alternating groups in intuitive terms? What to they do?

flag
5 
 You could probably ask this question on math.stackexchange.com; here it does not match the intended use of the site, which is research mathematics. – Mariano Suárez-Alvarez Sep 21 2010 at 16:33
4 
 A subgrup of order 6 would be normal with quotient of order 2, so abelian; it would then contain the derived subgroup $A_4'$ of $A_4$. But $A_4'$ has order $4$, and this is impossible. – Mariano Suárez-Alvarez Sep 21 2010 at 16:37
It might be worthwhile to point out that for $n\geq 5$, the alternating group $A_n$ on $n$ letters is a simple group; that is, it has no non-trivial proper normal subgroups. The alternating group $A_4$ has exactly one non-trivial proper normal subgroup (of order 4) - this normal subgroup is naturally isomorphic to the Klein 4-group. The alternating group $A_3$ is cyclic of order 3. There are many other important properties of alternating groups but I think these are the most basic. – Amitesh Datta Sep 22 2010 at 8:03

closed as too localized by Cam McLeman, Robin Chapman, Qiaochu Yuan, Peter Arndt, Noah Snyder Sep 21 2010 at 16:50

Browse other questions tagged or ask your own question.