Exercise 2.4.18 ($\{z \in \mathbb{C} \mid \exists n \in \mathbb{Z}^+,\ z^{p^n}= 1\}$ is not finitely generated)

Proof. Let $p$ be a prime and let

and for each $k\in {\mathbb{Z}}^{\text{+}}$ let

so that

(a)

• Suppose that $k\le m$. Let $z\in H_k$. Then $z\in Z$ and $z^{p^k}=1$. Therefore

  $$z^{p^m}={\left(z^{p^k}\right)}^{p^{m-k}}=1,$$

  where $m-k\ge 0$, thus $z\in H_m$. This shows that $H_k\le H_m$. So for all $k,m$ in ${\mathbb{Z}}^{\text{+}}$,

  $$k\le m\Rightarrow H_k\le H_m.$$

• Conversely, suppose that $H_k\le H_m$. Assume, for the sake of contradiction, that $m<k$. Put $z=e^{2\mathit{i\pi }∕p^{m+1}}$. Then $z^{p^m}=e^{2\mathit{i\pi }∕p}\ne 1$ (otherwise $1∕p\in \mathbb{Z}$ so $p\mid 1$: this is impossible because $p$ is a prime number). Therefore $z\notin H_m$. But, since $k-m-1\ge 0$

  $$z^{p^k}=e^{2\mathit{i\pi }{p}^{k-m-1}}=1,$$

  so $z\in H_k$, where $z\notin H_m$. This contradicts the inclusion $H_k\le H_m$. This contradiction proves that $k\le m$.

For all $k,m$ in ${\mathbb{Z}}^{\text{+}}$,

$$k\le m⟺H_k\le H_m.$$

(b)

We know by any course on complex numbers that for every $n\in {\mathbb{Z}}^{\text{+}}$, $$z^n=1⟺\exists <!--\; FUNCTION\; APPLICATION\; -->t\in \mathbb{Z},z=e^{2\mathit{\pi it}∕n}.$$

Therefore

$$\begin{array}{llll}\hfill z\in H_k& ⟺z^{p^k}=1& \hfill & \\ \hfill & ⟺\exists <!--\; FUNCTION\; APPLICATION\; -->t\in \mathbb{Z},z=e^{2\mathit{\pi it}∕p^k}& \hfill & \\ \hfill & ⟺\exists <!--\; FUNCTION\; APPLICATION\; -->t\in \mathbb{Z},z={\left(e^{2\mathit{\pi i}∕p^k}\right)}^t& \hfill & \\ \hfill & ⟺z\in ⟨e^{2\mathit{\pi i}∕p^k}⟩.& \hfill & \end{array}$$

Therefore

$$H_k=⟨e^{2\mathit{\pi i}∕p^k}⟩,$$

So $H_k$ is a cyclic group (of order $p^k$, since the order of $e^{2\mathit{\pi i}∕p^k}$ is $p^k$).

(c)

Note: If we want the trivial subgroup $\{1\}$ to be equal to some $H_k$, we must define $H_0=\{z\in Z\mid z^{p^0}=1\}=\{1\}$. So we take $k\in \mathbb{N}=\{0,1,2,3,\dots \}$ and not $k\in {\mathbb{Z}}^{\text{+}}=\{1,2,3,\dots \}$.

Let $H$ be any proper subgroup of $Z$. The order of every element of $Z$ is a power of $p$, so the order of every element of $H$ is a power of $p$.

Consider the set $A$ of integers $k$ such that $p^k$ is the order of some element in $H$:

$$A=\{k\in \mathbb{N}\mid \exists <!--\; FUNCTION\; APPLICATION\; -->h\in H,|h|=p^k\}.$$

(here $|h|$ is the order of $h$, and not its module!)

We want to prove that $A$ has a greatest element.

Since $1\in H$, where the order of $1$ is $1=p^0$, $0\in A$, so $A\ne \varnothing$.

Assume, for the sake of contradiction, that $A$ has no upper bound (see definition p. 908). Let $m$ be any element in ${\mathbb{Z}}^{\text{+}}$. We want to show that $H_m\le H$.

There is some $h_0\in H$ and $k_0\in \mathbb{N}$ such that $|h_0|=p^{k_0}>p^m$ (if all $h\in H$ satisfy $|h|\le p^m$ then $m$ is an upper bound for $A$). Then $h_0\in H_{k_0}$, and $|h_0|=p^{k_0}=|H_{k_0}|$, therefore $⟨h_0⟩=H_{k_0}\le H$. Since $H_m\le H_{k_0}$ by part (a), we obtain $H_m\le H$, for every $m\in {\mathbb{Z}}^{\text{+}}$, hence

$$Z=\underset{m\in {\mathbb{Z}}^{\text{+}}}{\bigcup }{H}_m\subseteq H,$$

where $H\le Z$, thus $H=Z$: this is false because by hypothesis $H$ is a proper subgroup of $Z$. This contradiction shows that $A$ is upper bounded (and not empty). Therefore there is a maximum $m$ in $A$:

$$m=\max <!--\; FUNCTION\; APPLICATION\; -->(A),$$

so that

• there is some $h\in H$ such that $|h|=p^m$, and
• for every $z\in H$, $|z|\le p^m$.

Since $|h|=p^m$, then as above $H_m=⟨h⟩\le H$. For every element $z\in H$, $z\in H_k$ for some integer $k$, where $|z|=p^k\le p^m$. By part (a), $H_k\le H_m$, so $z\in H_m$. Since this is true for every $z\in H$, we obtain $H\le H_m$, so $H=H_m$ (where $m=\max <!--\; FUNCTION\; APPLICATION\; -->(A)$).

In conclusion, every proper subgroup of $Z$ equals $H_m$ for some $m\in \mathbb{N}$ (in particular, every proper subgroup of $Z$ is finite and cyclic).

(d)

Assume that $Z$ is finitely generated, i.e. there are some elements $g_1,g_2,\dots ,g_n\in Z$ such that $$Z=⟨g_1,g_2,\dots ,g_n⟩.$$

Since $Z=\underset{k\in {\mathbb{Z}}^{\text{+}}}{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}{H}_k$, for every index $i\in [[1,n]]$, there is some $k_i$ such that

$$g_i\in H_{k_i}.$$

Put $k=\max <!--\; FUNCTION\; APPLICATION\; -->(k_1,k_2,\dots <!--\; FUNCTION\; APPLICATION\; -->,k_n)$. Then $k_i\le k$ for all $i$, thus $H_{k_i}\le H_k$ by part (a). Therefore $g_i\in H_k$ for all $i\in [[1,n]]$, so

$$Z=⟨g_1,g_2,\dots ,g_n⟩\le H_k.$$

Then for every integer $m\ge k$, $Z\le H_k\le H_m\le Z$, thus $Z=H_m$ for all $m\ge k$. In particular, $H_k=H_{k+1}$. This is impossible because by the converse part in part (a) $H_k<H_{k+1}$.

$Z$ is not finitely generated.