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Are the alternating groups $A_n$ nilpotent?

since $A_3$ are abelian so its clear that is nilpotent, but what about A_4? and in general A_n?

since finite group is nilpotent if and only if every maximal subgroup is normal. but after I looked at this paper which classify maximal subgroup of $A_n$ that we have a maximal subgroup but its clear that $A_n$ for $n\ge 5$ is simple ( have no normal ) which means $A_n$ is non-nilpotent for $n\ge 5$ . I want to know if that true and what about case 4. that $A_4$ is nilpotent?

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Every finite nonabelian group has (proper, nontrivial) maximal subgroups: take any cyclic subgroup generated by a nontrivial element (necessarily proper), then just keep enlargening the subgroup (strictly within the overgroup) until you can't any more (this process necessarily terminates because the group is finite). Thus, there is no need to classify maximal subgroups of the simple alternating groups to see that they are not nilpotent.

If you google alternating group A4 and go to the GroupProps page and head down to the table in the "group properties" section, you will see it says it is not nilpotent, with the explanation on the side that $A_4$ is centerless (and thus, has no ascending central series). This can be verified directly by hand (it suffices to verify noncentrality of a set of representatives for conjugacy classes, which corresponds to verifying noncentrality of one permutation per nontrivial cycle type, of which there are only two).

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A finite group is nilpotent if and only if it is a direct product of its Sylow subgroups. Since $A_n$ is simple for $n>4$, this clearly cannot happen.

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    $\begingroup$ It's much easier than that: $A_n$ has trivial abelianization for $n\ge 5$ (easier to prove than simplicity), and no such group can even be solvable, let alone nilpotent. $\endgroup$ Oct 1 '15 at 19:38
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    $\begingroup$ @QiaochuYuan Yes, I wasn't aiming for easy. I wanted to state a fact that is not that trivial and sheds some light on the structure of finite nilpotent groups. $\endgroup$
    – Pedro Tamaroff
    Oct 2 '15 at 0:18

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