Quotient of nilpotent group is nilpotent

Question

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Edit: I managed to rephrase my proof in a way that does not resort to coset multiplication. I think the resulting proof is better. I've added it as an answer below, while preserving the original question to avoid wrecking the context.

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In his book Finite Group Theory, section 1.D, Isaacs mentions without proof the following result:

Proposition 1: If $G$ is a nilpotent group and $H \lhd G$, then $G/H$ is nilpotent.

I think I have a proof but would appreciate confirmation and any suggestions for improvement.

Isaacs' definition is that $G$ is nilpotent if it has a central series:

$G$ is nilpotent if it has a central series, meaning that there are subgroups $N_i \lhd G$ such that $1 = N_0 \leq N_1 \leq \cdots \leq N_r = G$ and $N_i / N_{i-1} \leq Z(G / N_{i-1})$.

I would like to work directly with this definition rather than with upper or lower central series. The reasons for this: (1) Isaacs' treatment of upper central series assumes that the group is finite, and uses the result I am trying to prove in order to establish the equivalence between "$G$ has a central series" and "$G$ has an upper central series which reaches $G$"; (2) lower central series are not covered until a later chapter.

Initially I tried to prove this by defining $M_i = N_i H / H$ and attempting to show that this results in a central series. I think I made it work using a somewhat messy argument involving the isomorphism and correspondence theorems, but then it occurred to me that instead it suffices to prove the following result:

Proposition 2: If $G$ is nilpotent and $\phi : G \to H$ is an epimorphism, then $H$ is nilpotent.

Then we can apply this to the canonical epimorphism $\phi : G \to G/H$ given by $\phi(g) = gH$, which gives the desired result.

Proof: We have $N_i \lhd G$ and $1 = N_0 \leq N_1 \leq \cdots \leq N_r = G$, with $N_i / N_{i-1} \leq Z(G / N_{i-1})$. The latter condition can be stated equivalently as follows: if $n \in N_i$ and $g \in G$, then $ngN_{i-1} = gnN_{i-1}$.

Now define $M_i = \phi(N_i)$ for every $0 \leq i \leq r$. Note that $M_0 = \phi(N_0) = \phi(1) = 1$, and $M_r = \phi(N_r) = \phi(G) = H$. Also, $M_i \lhd H$ by the correspondence theorem since $N_i \lhd G$.

Now, let $m \in M_i$ and $h \in H$. To show that $M_i / M_{i-1} \leq Z(H / M_{i-1})$, it suffices to show that $mhM_{i-1} = hmM_{i-1}$. We have $m = \phi(n)$ and $h = \phi(g)$ for some $n \in N_i$ and $g \in G$. Then $mhM_{i-1} = \phi(ngN_{i-1}) = \phi(gnN_{i-1}) = hmM_{i-1}$, where the second equality holds because $N_{i} / N_{i-1} \leq Z(G / N_{i-1})$. We conclude that $1 = M_0 \leq M_1 \leq \cdots\leq M_r = H$ is a central series for $H$, so $H$ is nilpotent.

Answer 1

I cleaned up the proof so that it no longer uses the somewhat unprofessional "coset multiplication" argument. I'm happy with it now, so I'll post it as an answer in case it is of interest to anyone.

Recall that the goal is to prove the following:

If $G$ is nilpotent and $N \lhd G$, then $G/N$ is nilpotent.

First, we prove a simple lemma:

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Lemma: If $\theta : G \to H$ is an epimorphism, then $\theta(Z(G)) \leq Z(H)$.

Proof: Let $z \in Z(G)$ and $h \in H$. Then $h = \theta(g)$ for some $g \in G$, and $\theta(z)h = \theta(zg) = \theta(gz) = h\theta(z)$. Therefore, $\theta(z) \in Z(H)$.

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Now we prove the following proposition:

Proposition: If $G$ is nilpotent and $\phi : G \to H$ is an epimorphism, then $H$ is nilpotent. (Note: Applying this proposition to the canonical epimorphism $G \to G/N$ will yield the desired result.)

Proof: Since $G$ is nilpotent, we have $N_i \lhd G$ and $1 = N_0 \leq N_1 \leq \cdots \leq N_r = G$, with $N_i / N_{i-1} \leq Z(G / N_{i-1})$.

Define $M_i = \phi(N_i)$. We claim that $1 = M_0 \leq M_1 \leq \cdots \leq M_r = H$ is a central series for $H$.

First, we observe that $M_0 = \phi(N_0) = \phi(1) = 1$ and $M_r = \phi(N_r) = \phi(G) = H$. Also, $M_i \lhd H$ by the correspondence theorem, since $N_i \lhd G$.

Now define $\theta : G / N_{i-1} \to H / M_{i-1}$ by $\theta(gN_{i-1}) = \phi(g)M_{i-1}$. It is routine to verify that $\theta$ is a well-defined epimorphism. We can therefore apply the lemma to conclude that $\theta(Z(G/N_{i-1})) \leq Z(H / M_{i-1})$.

Note also that $\theta(N_i / N_{i-1}) = \phi(N_i) / M_{i-1} = M_i / M_{i-1}$.

Since $G$ is nilpotent, we have $N_i / N_{i-1} \leq Z(G/N_{i-1})$. Consequently, $M_i / M_{i-1} = \theta(N_i / N_{i-1}) \leq \theta(Z(G/N_{i-1})) \leq Z(H / M_{i-1})$, as required.

Answer 2

Your argument is correct. A simpler and better way is as follows: $$N_i/N_{i-1} \leq Z(G/N_{i-1}) \Longleftrightarrow N_i/N_{i-1} \mbox{ commutes with } G/N_{i-1} \Longleftrightarrow [N_i/N_{i-1}, G/N_{i-1}]=1 \Longleftrightarrow [N_i,G]\leq N_{i-1} $$ By homomorphism $\phi$, we have $\phi[x,y]=[\phi(x),\phi(y)]$. Hence, $$ [M_i, H]=[\phi(N_i), \phi(G)]=\phi[N_i,G]\leq \phi(N_{i-1})=M_{i-1}.$$ This implies $H$ is nilpotent with central series $\{M_i\}_i$.

Answer 3

I am indebted to Not Euler's answer and provide the slightly more general result for both subgroups and quotient groups of a nilpotent group.

Lemma 1. $g\in Z_{n}\left(G\right)$ if and only if $gxg^{-1}x^{-1}\in Z_{n-1}\left(G\right)$ for all $x\in G$.

Proof of Lemma 1. The result follows immediately from the definition \begin{align*} Z_{n}\left(G\right)/Z_{n-1}\left(G\right) &=Z\left(G/Z_{n-1}\left(G\right)\right) \end{align*}

Lemma 2. Let $H$ be any subgroup of $G$. Then $H\cap Z_{n}\left(G\right)\leq Z_{n}\left(H\right)$.

Proof of Lemma 2. Induct on the index $n$. It is obvious that $H\cap Z\left(G\right)\leq Z\left(H\right)$ so assume the induction hypothesis and consider $h\in H\cap Z_{n+1}\left(G\right)$ so there is the sequence of implications \begin{align*} &\phantom{\Rightarrow\ \;}h\in Z_{n+1}\left(G\right)\\ &\Rightarrow hxh^{-1}x^{-1}\in Z_{n}\left(G\right)\text{ for all x}\in G\\ &\Rightarrow hxh^{-1}x^{-1}\in H\cap Z_{n}\left(G\right)\text{ for all x}\in H \end{align*} and the induction hypothesis implies $hxh^{-1}x^{-1}\in Z_{n}\left(H\right)\text{ for all x}\in H$ so, by Lemma~1, $h\in Z_{n+1}\left(H\right)$. In other words, $H\cap Z_{n+1}\left(G\right)\leq Z_{n+1}\left(H\right)$. Thus Lemma 2.

Lemma 3. Let $H$ be a normal subgroup of $G$. Then $HZ_{n}\left(G\right)/H\leq Z_{n}\left(G/H\right)$.

Proof of Lemma 3. Induct on the index $n$. For $n=0$, it is trivial that $H\leq H$. It is satisfying and canonical to prove the non-trivial base step $HZ\left(G\right)/H\leq Z\left(G/H\right)$. Consider $hgH\in HZ\left(G\right)/H$. Then $g\in Z\left(G\right)$ so, by Lemma 1, $gxg^{-1}x^{-1}=1\in Z_{0}\left(G\right)$ for all $x\in G$. Trivially, or by the induction hypothesis, $gxg^{-1}x^{-1}H=H\in Z_{0}\left(G/H\right)$ for all $xH\in G/H$ so, by Lemma 1, $gH\in Z\left(G/H\right)$. In other words, $hgH=gh^{\prime}H=gH\in Z\left(G/H\right)$ so $HZ\left(G\right)/H\leq Z\left(G/H\right)$. The general step of the induction is structured similarly. Thus Lemma 3.

Proof. For the first result, let $H$ be any subgroup of $G$ and let $n$ be the nilpotence class of $G$. Then Lemma 2 implies $H$ has a nilpotence class less than or equal to $n$. That is, $H$ is nilpotent.

For the second result, let $H$ be a normal subgroup of $G$ and let $n$ be the nilpotence class of $G$. Then Lemma 3 implies $G/H$ has a nilpotence class less than or equal to $n$. That is, $G/H$ is nilpotent.

Answer 4

Hint: Suppose $1\triangleleft Z_1(G)\triangleleft Z_2(G)\triangleleft ... \triangleleft Z_m(G)=G$. Let $N\triangleleft G$. Notice that $Z_m(G)N/N=G/N$. So if we could show that $Z_m(G)N/N\subset Z_m(G/N)$ we are done. Just use induction to prove this fact. It is very straightforward.