Need help proving any subgroup and quotient of a nilpotent group is nilpotent?

Question

$G$ is the direct product of its Sylow subgroups $P_i$. Then if $H \le G$, $P_i \cap H \le H$. I'm stuck now on how to proceede to find a terminating central series for $H$.

Also, I know $P_i H / H \le G/H$ so I'm assuming proving the above will prove that the quotient is nilpotent too?

Answer 1

I posted a proof that a quotient of a nilpotent group is nilpotent recently.

Here is a proof that a subgroup of a nilpotent group is nilpotent.

I will work with the definition that a group $G$ is nilpotent if it has a central series $$1 = N_0 \leq N_1 \leq \cdots \leq N_r = G$$ where each $N_i \lhd G$ and $N_i / N_{i-1} \leq Z(G/N_{i-1})$ for each $i$.

Suppose that $G$ is nilpotent and $H \leq G$. We wish to show that $H$ is nilpotent. Define $M_i = H \cap N_i$ for $0 \leq i \leq r$. Note that $M_0 = H \cap N_0 = H \cap 1 = 1$, and $M_r = H \cap N_r = H \cap G = H$. Also, $H \cap N_{i} \lhd H$ by the diamond isomorphism theorem.

It remains to show that $$M_i / M_{i-1} \leq Z(H / M_{i-1})$$

Once again by the diamond isomorphism theorem, we have an isomorphism $\phi : H / M_{i-1} \to HN_{i-1}/N_{i-1}$ given by $\phi(hM_{i-1}) = hN_{i-1}$. The image of $M_i / M_{i-1}$ is $\phi(M_i / M_{i-1}) = M_{i}N_{i-1}/N_{i-1}$.

Since $M_{i} = H \cap N_{i} \leq N_{i}$ and $N_{i-1} \leq N_{i}$, it follows that $M_{i}N_{i-1}/N_{i-1} \leq N_i / N_{i-1}$, so $\phi(M_{i} / M_{i-1}) \leq N_i / N_{i-1} \leq Z(G/N_{i-1})$.

Therefore, $$\begin{aligned} \phi(M_i/M_{i-1}) &\leq (HN_{i-1} / N_{i-1}) \cap Z(G/N_{i-1}) \\ &\leq Z(HN_{i-1} / N_{i-1})\\ & = Z(\phi(H/M_{i-1}))\\ &= \phi(Z(H/M_{i-1}))\\ \end{aligned}$$ where the last equality follows because $\phi$ is an isomorphism, so the center is preserved by $\phi$.

Finally, again since $\phi$ is an isomorphism, we can apply $\phi^{-1}$ to both sides to conclude that $M_i / M_{i-1} \leq Z(H/M_{i-1})$, as desired.

Answer 2

Here is a completely different proof which uses the characterization that if $G$ is finite, then the following conditions are equivalent:

1. $G$ is nilpotent (it has a central series from $1$ to $G$)
2. For every $p$ dividing $G$, there is a unique (hence normal) Sylow $p$-subgroup.
3. $G$ is the direct product of its nontrivial Sylow subgroups.

The equivalence of 2 and 3 is easy to verify: if $G$ is a direct product of a collection of subgroups, then by definition those subgruops are normal in $G$. Conversely, if every Sylow $p$-subgroup of $G$ is normal, then we can form their product, say $S_1 S_2 \cdots S_n$. By computing the order of the product we see that it is all of $G$, and each $S_i$ intersects trivially with $\prod_{j\neq i}S_j$, so the product is direct.

Let us use characterization 2: assume that each Sylow $p$-subgroup of $G$ is unique.

Let $H \leq G$, and let $S \in Syl_p(G)$. Then $P = S \cap H$ is a $p$-subgroup of $H$, so it is contained in some $T \in Syl_p(H)$. It is easy to show that $T$ must be of the form $R \cap H$ for some $R \in Syl_p(G)$. But there is only one Sylow $p$-subgroup of $G$, so $R = S$, which means that $T = S \cap H = P$, hence $P \in Syl_p(H)$.

Therefore, if $S_1,\ldots,S_n$ are the nontrivial Sylow subgroups of $G$, and we set $P_i = S_i \cap H$, then $P_1,\ldots,P_n$ includes all of the nontrivial Sylow subgroups of $H$. Some of the $P_i$ may be trivial, but that doesn't affect the rest of the proof.

It remains to show that each $P_i$ is normal in $H$. But this follows immediately from the diamond isomorphism theorem: $P_i = S_i \cap H \lhd H$ because $S_i \lhd S_iH$.

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Caveats:

1. This proof is valid only if $G$ is finite.
2. If you use this proof, be sure to check how you established the equivalence of 1 and 2 (or equivalently, 1 and 3). If your proof of that equivalence used the fact that a subgroup of a nilpotent group is nilpotent, then this proof results in a circular argument.

Answer 3

Let $H$ be a subgroup of the nilpotent group $G$. Pick a central series $$ \langle1\rangle = N_0\subseteq N_1\subseteq\cdots\subseteq N_r=G. $$ Then $$ \langle1\rangle=H\cap N_0\subseteq H\cap N_1\subseteq\cdots\subseteq H\cap N_r=H $$ is a normal series in $H$ because $H\cap N_i\triangleleft H$. Moreover, for every $y\in H$ and $z\in H\cap N_i$ we have $$ \text{ad}_y(z)=yzy^{-1}z^{-1}\in H\cap N_{i-1}. $$ Thus, $\text{ad}_y(H\cap N_i)\subseteq H\cap N_{i-1}$, i.e., the series for $H$ is central.