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If $G$ is a group and $N<Z(G)$ with $G/N$ nilpotent then I want to show that $G$ is also nilpotent (here we take the definition of nilpotency for finite groups to be has a unique sylow subgroup of each order).

My proof of this claim is below, however this result seems similar to other results (about soluble groups and so on) where the proofs are a lot faster. Have I over complicated this (gotten it wrong?) or is there a faster way to do this? (possibly with a definition of nilpotency generalized to infinite groups-which my lecturer said that there was?) Also is there a more general result with $N$ just having to be nilpotent rather than $N\leq Z(G)$

So we have the canonical homomorphism $f:G\rightarrow G/N$. First we show that any Sylow-subgroup gets mapped to a Sylow subgroup of $G/N$. Take $P\in Syl_p(G)$ then we have that $f(P)=PN/N\cong P/(P\cap N)$ by (one of) the isomorphism theorems.

Then by an order argument $|P|=p^n$ and so $|P\cap N|=p^m$ for some $ m\leq n$ and $p^m||N|$ maximally for a power of $p$ and hence this is also a $p$-group.

Now in order for $G$ not to be nilpotent we need that there are more than one Sylow p-group for some p, say $P$ and $Q$.

Now if we suppose that was the case then we would have that $f(P)=f(Q)$. So suppose that we do.

Then I think that I want to show that $P$ and $Q$ have to be conjugate by an element in $N$, then I would be done.

Claim $P$ and $Q$ are conjugate by an element of $N$

So if $f(P)=f(Q)$ then $f^{-1}f(P)=f^{-1}f(Q)$ which gives that:

$NP=NQ$

Now $P$ is a sylow group of $NP$ and $Q$ is a sylow group of $NQ$ (just by definition as they are both in the group and are p-sylow groups of the bigger group $G$)

We then know that all sylow p-groups are conjugate and so we have that $\exists x\in NP$ such that $x P x^{-1}= Q$ (1)

But then as $x\in NP$ we have that $x=np$ for some $n\in N$ and $p\in P$ and so this gives us from (1) that:

$npPp^{-1}n=nPn^{-1} =Q$ and hence claim.

We then have that $P=nQn^{-1}$ but as $n\in N$ then this gives that $nQn^{-1}=Q$ and so we have that $P=Q$ and hence $G$ has unique Sylow subgroups of each order and we are done

hmmmm
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1 Answers1

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@hmmmm: The general direction of your argument is correct, but you can make use of a few auxiliary results to shorten it a bit:

Let $P$ be a Sylow-$p$-subgroup of $G$ and set $H:=PN$.

Since $f(H)$ is a (the) Sylow-$p$-subgroup of the nilpotent group $f(G)$, $H$ must be normal in $G$, so we get

$$\begin{align} G&=N_G(P)H & \text{(this is the Frattini argument)}\\ &=N_G(P)PN \\ &=N_G(P)N & \text{(since $P\leq N_G(P)$)}\\ &=N_G(P)Z(G) & \text{(since $N\leq Z(G)$)}\\ &=N_G(P) & \text{(since $Z(G) \leq N_G(P)$),} \end{align}$$

so $P$ is normal in $G$ and thus the only Sylow-$p$-subgroup (since by Sylow's Theorem all such Sylow-$p$-subgroups are conjugate).

jpvee
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