Prove $G$, $|G|=n$ is nilpotent $\iff$ $\forall m|n$, $G$ has a normal subgroup of order $m$. I got stuck in the second direction.

One direction: $|G|=n=p_1^{s_1}\cdot ...\cdot p_k^{s_k}$ Where $p_i$ prime. Particularly, $\forall p_i^{s_i}, p_i^{s_i}|n$ and therefore the Sylow-$p_i$ subgroup is unique and normal. Therefore every Sylow-$p$ subgroup is normal and that means $G$ is nilpotent. (There is a theorem\corollary claiming that saying every Sylow-$p$ subgroup is normal is equivalent to saying $G$ is nilpotent.

Other direction: Let $G$ be nilpotent. $|G|=n=p_1^{s_1}\cdot ...\cdot p_k1^{s_k}$ Where $p_i$ prime. Then, Sylow-$p_i$ subgroup is unique and normal. But what about the $p$-subgroups of order such as $p_i^{},p_i^{2},...,p_i^{s_i-1}$? They are subgroups contained in the Sylow-$p_i$ subgroup, and they all divide $n$, but are they normal? How can I show that?

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    $\begingroup$ Not all such subgroups are normal, but you only need to show that for every group $P$ of order $p^k$ (with $p$ prime), and every $j \le k$, $P$ has a normal subgroup of order $p^j$. You could do this by induction on $j$, using the fact that $Z(P) \ne 1$ for the base case $j=1$. $\endgroup$ – Derek Holt Jan 21 '15 at 11:53
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    $\begingroup$ That might be a characteristic subgroup of a normal subgroup, which makes a world of difference. $\endgroup$ – Jyrki Lahtonen Jan 21 '15 at 12:02
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    $\begingroup$ Even outer automorphisms of $P$ map $Z(P)$ to itself. $\endgroup$ – Jyrki Lahtonen Jan 21 '15 at 12:03
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    $\begingroup$ You don't need to worry about characteristic subgroups - that's a red herring! A finite nilpotent group is the direct product $P_1 \times \cdots \times P_k$ of its Sylow subgroups, so any normal subgroup of any $P_i$ is also normal in $G$. $\endgroup$ – Derek Holt Jan 21 '15 at 12:52
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    $\begingroup$ Meitar, in an (inner) direct product of two subgroups $G=H\times K$ we have that 1) all the elements $g$ of $G$ can be uniquely written in the form $g=hk$ with $h\in H, k\in K$. AND 2) $hk=kh$ for all $h\in H, k\in K$. Now if you take a normal subgroup of $H$ and conjugate it with $g$,... $\endgroup$ – Jyrki Lahtonen Jan 21 '15 at 14:02


Since $\;G\;$ is nilpotent then every Sylow subgroup is normal. Now use the following:

Lemma 1: If $\;H\;$ is a finite $\;p$- group, $\;|H|=p^n\;$ , then for all $\;0\le k\le n\;$ there exists a normal subgroup $\;K\;$ of $\;H\;$ of order $\;p^k\;$

Proof (hints): use that $\;Z(H)\neq 1\;$, induction + the correspondence theorem (a subgroup of a quotient group is normal iff its inverse image under the quotient homomorphism is normal in the whole group)

Finally, as $\;G\;$ is the direct product of its Sylow subgroups and each pair of these commutes (why?) and etc.

Using the above (or whatever), you can also try the following nice

Lemma: The group $\;G\;$ is nilpotent iff $\;xy=yx\;$ for any pair of elements in the group with coprime orders.

  • $\begingroup$ Thank you :) The characteristic thing is what I was looking for :) $\endgroup$ – Meitar Jan 21 '15 at 12:38
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    $\begingroup$ I'm afraid the argument in the end is false. The correct logic is: a characteristic subgroup of a normal subgroup is itself normal. You are saying: a normal subgroup of a characteristic subgroup is normal. This is not always true. $\endgroup$ – Jyrki Lahtonen Jan 21 '15 at 12:39
  • $\begingroup$ To fix things in the end you need to use the fact that a nilpotent is a direct product of its Sylow subgroups. In a direct product elements of various factors commute, so... $\endgroup$ – Jyrki Lahtonen Jan 21 '15 at 12:46
  • $\begingroup$ Yes the characteristic thing is problematic... I remember proving the lemma... I sweat blood $\endgroup$ – Meitar Jan 21 '15 at 13:52
  • $\begingroup$ Not only it is false: it is even not true :) .I'm editing the answer, and thanks to Jirky. $\endgroup$ – Timbuc Jan 21 '15 at 14:05

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