## Basic properties of finite nilpotent groups

Let $G$ be a finite nilpotent group.

1. If $H$ is a nontrivial normal subgroup of $G$, then $H$ intersects the center of $G$ nontrivially. In particular, every normal subgroup of prime order is central.
2. If $H < G$ is a proper subgroup then $H$ is properly contained in $N_G(H)$.

First we prove some lemmas.

Lemma 1: Let $G$ be a group, with $A_1,A_2,B_1,B_2 \subseteq G$. If $A_1 \subseteq A_2$ and $B_1 \subseteq B_2$, then $[A_1,B_1] \leq [A_2,B_2]$. Proof: $[A_1,B_1]$ is generated by $[a,b]$ such that $a \in A_1$ and $b \in B_1$. Each of these generators is in $[A_2,B_2]$. $\square$

Lemma 2: Let $G$ be a group. If $H \leq G$, then $H^k \leq G^k$ for all $k$, where $G^k$ denotes the $k$-th term in the lower central series of $G$. Proof: We proceed by induction on $k$. For the base case, if $k=0$ then $H^0 \leq G^0$. For the inductive step, suppose $H^k \leq G^k$ for some $k \geq 0$. Then by Lemma 1, $H^{k+1} = [H,H^k] \leq [G,G^k] = G^{k+1}$. $\square$

Lemma 3: If $G$ is a nilpotent group of nilpotence class $k$ and $H \leq G$, then $H$ is nilpotent with nilpotence class at most $k$. Proof: By Theorem 8 in the text, $G^k = 1$. By Lemma 2, $H^k = 1$, so that $H$ is nilpotent and of nilpotence class at most $k$. $\square$

Lemma 4: Let $\varphi : G \rightarrow H$ be a group homomorphism and let $S \subseteq G$. Then $\varphi[\langle S \rangle] = \langle \varphi[S] \rangle$. Proof: We have $x \in \varphi[\langle S \rangle]$ if and only if $x = \varphi(s_1^{a_1} \cdots s_k^{a_k})$ for some $s_i \in S$ if and only if $x = \varphi(s_1)^{a_1} \cdots \varphi(s_k)^{a_k}$ for some $s_i \in S$ if and only if $x \in \langle \varphi[S] \rangle$. $\square$

Lemma 5: Let $\varphi : G \rightarrow H$ be a surjective group homomorphism. Then $H^k \leq \varphi[G^k]$ for all $k$, where $G^k$ denotes the $k$-th term in the lower central series of $G$. Proof: We proceed by induction on $k$. For the base case, if $k = 0$ then $H^0 = H = \varphi[G] = \varphi[G^0]$. For the inductive step, suppose $H^k \leq \varphi[G^k]$. Then we have $H^{k+1} = [H,H^k] \leq [\varphi[G],\varphi[G^k]]$ $\leq \varphi[[G,G^k]]$ $\leq \varphi[G^{k+1}]$. $\square$

Lemma 6: Let $G$ be a nilpotent group of nilpotence class $k$. If $H \leq G$ is normal, then $G/H$ is nilpotent of nilpotence class at most $k$. Proof: By Lemma 5, we have $(G/H)^k \leq \varphi[G^k] = 1$. $\square$

Now to the main results; first we show part 2. [With hints from MathReference.]

(2) Let $G$ be a finite nilpotent group such that $Z(G) \neq 1$. If $H \leq G$ is a nontrivial normal subgroup of $G$, then $H \cap Z(G)$ is nontrivial. In particular, every normal subgroup of order $p$ is contained in the center.

We proceed by induction on the breadth of $G$. (Recall: the breadth of a finite group is the number of prime factors dividing its order, including multiplicity.)

For the base case, if $G$ has breadth 1, then $G \cong Z_p$ for some prime $p$ and thus $G$ is simple and abelian. Thus $H = G$, and $H \cap Z(G) \neq 1$.

For the inductive step, suppose that the conclusion holds for every finite nilpotent group of breadth at most $k \geq 1$, and let $G$ be a finite nilpotent group of breadth $k+1$. Let $H \leq G$ be normal and suppose $H \cap Z(G) = 1$. Now $H \cdot Z(G) \cong H \times Z(G) \leq G$ is an internal direct product. Moreover, since $H$ and $Z(G)$ are normal in $G$, $H \cdot Z(G) \leq G$ is normal. Now by the Lattice Isomorphism Theorem, $HZ(G)/Z(G) \leq G/Z(G)$ is normal. By Lemma 6, $G/Z(G)$ is nilpotent, has breadth at most $k$, and $HZ(G)/Z(G) \leq G/Z(G)$ is normal. By the inductive hypothesis, there exists an element $xZ(G) \in Z(G/Z(G))$ such that $xZ(G) \in HZ(G)/Z(G)$. We can write $x = hz$, where $h \in H$ and $z \in Z(G)$, so that in fact $hZ(G) \in Z(G/Z(G))$. Now let $g \in G$ be arbitrary; since $H$ is normal in $G$, $ghg^{-1} \in H$. Now $(ghg^{-1})Z(G) = (gZ(G))(hZ(G))(g^{-1}Z(G))$ $= hZ(G)$ since $hZ(G)$ is in the center of $G/Z(G)$, so that $ghg^{-1}Z(G) = hZ(G)$. Write $ghg^{-1} = hk$, where $k \in Z(G)$. Now $ghg^{-1}h^{-1} = k \in Z(G)$, and since $H \cap Z(G) = 1$, we have $ghg^{-1}h^{-1} = 1$, thus $gh = hg$. So $h \in H$ is in the center of $G$ and $h \neq 1$, a contradiction. Thus $H \cap Z(G)$ is nontrivial. $\square$

Next, we prove some more lemmas.

Lemma 7: Let $G$ be a group, $K, H_1, H_2 \leq G$ be subgroups with $K \leq G$ normal, $K \leq H_1$, and $K \leq H_2$. If $H_1/K = H_2/K$, then $H_1 = H_2$. Proof: Let $x \in H_1$. Then $xK \in H_1/K = H_2/K$, so that $xK = yK$. In particular, $x \in yK \leq H_2$. The other direction is similar. $\square$

Lemma 8: Let $G$ be a group, $H \leq G$ a normal subgroup, and $K$ a subgroup with $H \leq K \leq G$. Then $N_{G/H}(K/H) = N_G(K)/H$. Proof: Note that $xH \in N_{G/H}(K/H)$ if and only if $(xH)(K/H)(x^{-1}H) = K/H$, if and only if $(xKx^{-1})/H = K/H$, if and only if (by Lemma 7) $xKx^{-1} = K$, if and only if $x \in N_G(K)$, if and only if $xH \in N_G(K)/H$. $\square$

Now we show part 4.

(4) Let $G$ be a finite nilpotent group. If $H < G$ is a proper subgroup, then $H < N_G(H)$ is proper.

Proof: We proceed again by induction on the breadth of $G$.

For the base case, if $G$ has breadth 1, then $G \cong Z_p$ is abelian. Then if $H \leq G$ is proper, $H \leq N_G(H) = G$ is proper.

For the inductive step, suppose the conclusion holds for all finite nilpotent groups of breadth at most $k$. Let $G$ be a finite nilpotent group of breadth $k+1$ and let $H \leq G$ be a proper subgroup. Now $Z(G) \leq N_G(H)$. If $Z(G) \not\leq H$, then $H \leq \langle H, Z(G) \rangle$ is proper, so that $H \leq N_G(H)$ is proper. If $Z(G) \leq H$, then $H/Z(G) \leq G/Z(G)$ is proper. Moreover, $G/Z(G)$ is a finite nilpotent group of breadth at most $k$, so that $H/Z(G) \leq N_{G/Z(G)}(H/Z(G))$ is proper. By Lemma 8, $H/Z(G) \leq N_G(H)/Z(G)$ is proper, so that $H \leq N_G(H)$ is proper. $\square$