## Equivalent characterizations of nilpotent and cyclic groups

If $G$ is finite, prove that (1) $G$ is nilpotent if and only if it has a normal subgroup of each order dividing $|G|$, and (2) is cyclic if and only if it has a unique subgroup of each order dividing $|G|$.

We begin with some lemmas.

Lemma 1: Let $G$ be a group, and let $H_1, \ldots, H_k$ be characteristic subgroups of $G$. Then $H = H_1\cdots H_k$ is characteristic in $G$. Proof: $H$ is a subgroup because each $H_i$ is normal. If $\alpha$ is an automorphism of $G$, then $\alpha[H] = \alpha[H_1] \cdots \alpha[H_k]$ $= H_1 \cdots H_k = H$. $\square$

Lemma 2: Let $G$ be a finite group. If $P \leq G$ is a normal Sylow subgroup, then $P$ is characteristic. Proof: $P$ is the unique subgroup of order $|P|$. $\square$

Now to the main result; first we prove (1).

$(\Rightarrow)$ We proceed by induction on the breadth (number of prime divisors with multiplicity) of $G$.

For the base case, if $G$ is a finite nilpotent group of breadth 1, then $G \cong Z_p$ is an abelian simple group. Thus $G$ has a normal subgroup of order 1 and $p$, the only divisors of $|G|$.

For the inductive step, suppose that every finite nilpotent group of breadth $k \geq 1$ has a normal subgroup of each order dividing $|G|$, and let $G$ be a finite nilpotent group of breadth $k+1$. Now $Z(G)$ is nontrivial, so there exists an element $x \in Z(G)$ of prime order, and $\langle x \rangle G$ is normal. We showed in a lemma to the previous exercise that $G/\langle x \rangle$ is nilpotent, and has breadth $k$. By the inductive hypothesis, $G/\langle x \rangle$ has a normal subgroup of each order $d$ dividing $|G/\langle x \rangle|$. By the Lattice Isomorphism Theorem, a subgroup $\overline{H} \leq G/\langle x \rangle$ is normal if and only if $\overline{H} = H/\langle x \rangle$ for some normal subgroup $H \leq G$. Thus we see that $G$ has a normal subgroup of order $d$ for all $d$ dividing $|G|$ such that $p|d$. Now let $d$ divide $|G|$, with $p$ not dividing $d$, and let $H \leq G$ be a normal subgroup of order $pd$. Since $H \leq G$, $H$ is a finite nilpotent group. By Theorem 3, $H$ is the direct product of its Sylow subgroups. Write $H = P \times Q$, where $P$ is the Sylow $p$ subgroup of $H$ and $Q$ the direct product of the remaining Sylow subgroups. By Lemma 2, each factor of $Q$ is characteristic in $H$, so that $Q$ is characteristic in $H$. Note that $|Q| = d$, thus $Q \leq G$ is normal and has order $d$.

$(\Leftarrow)$ Suppose $G$ has a normal subgroup of each order $d$ dividing $|G|$. In particular, $G$ has a normal (thus unique) Sylow $p$-subgroup for each prime $p$. By Theorem 6.3, $G$ is nilpotent.

Now we prove (2).

The $(\Rightarrow)$ direction was proved in Theorem 2.7.

$(\Leftarrow)$ Suppose $G$ is finite and has a unique subgroup of order $d$ for each $d$ properly dividing $|G|$. In particular, all the Sylow subgroups of $G$ are unique, hence normal; say there are $k$ of these. Suppose some Sylow $p$-subgroup $P$ of $G$ is not cyclic; then by this previous exercise, $P$ has a subgroup congruent to $Z_p \times Z_p$, and thus $G$ has distinct subgroups of order $p$, a contradiction. Thus each Sylow subgroup of $G$ is cyclic; say $P_i = \langle x_i \rangle$. Then $|x_1 \cdots x_k| = |G|$, so that $G = \langle x_1 \cdots x_k \rangle$ is cyclic.