Exercise 2.4.19 (No finite abelian group is divisible)

Notations: $a\wedge b=g.c.d.(a,b),a\vee b=l.c.m.(a,b)$.

(a)

Let $a\in \mathbb{Q}$, and let $k$ be a nonzero integer. Then $a=\frac{p}{q}$, where $p,q$ are integers, $q\ne 0$. Put $x=\frac{p}{\mathit{qk}}$. Then $a=\mathit{kx}$. This shows that $\mathbb{Q}$ is divisible.

(b)

Let $(G,\cdot )$ be an abelian group. Let $e$ be the least common multiple of the orders of the elements ( $e$ is the exponent of the group $G$). Then , for all $x\in G$, the order of $x$ divides $e$, so $x^e=1$. Let us denote $\omega (G)$ the exponent of $G$.

We want to show that there is some element $a\in G$ such that $|a|=e=\omega (G)$ (Source: Demazure “Cours d’algèbre”: I detailed the arguments.)

Lemma 1. Let $x$ and $y$ be two elements of finite order in a group $G$ such that $\mathit{xy}=\mathit{yx}$. If $|x|=n$ and $|y|=m$, where $m\wedge n=1$, then $|\mathit{xy}|=\mathit{mn}$.

Proof. First, since $\mathit{xy}=\mathit{yx}$, then ${(\mathit{xy})}^{\mathit{mn}}={(x^n)}^m{(x^m)}^n=1$. Therefore, for all integers $k$, if $\mathit{mn}\mid k$, then ${(\mathit{xy})}^k=1$.

Conversely, let $k$ be an integer such that ${(\mathit{xy})}^k=1$. Then $1={(\mathit{xy})}^{\mathit{kn}}={(x^n)}^k(y^{k}n)=y^{\mathit{kn}}$. Therefore $m\mid \mathit{kn}$, where $m\wedge n=1$, so $m\mid k$. Similarly $n\mid k$. Therefore $m\vee n\mid k$, and since $m\wedge n=1$, $m\vee n=\mathit{mn}$, so $\mathit{mn}\mid k$. This shows that for all integers $k$,

$${(\mathit{xy})}^k=1⟺\mathit{mn}\mid k.$$

Hence $|\mathit{xy}|=\mathit{mn}$. □

Lemma 2. Let $G$ be a finite abelian group, and $a,b\in G$. There exists some element $c\in G$ such that $|c|=|a|\vee |b|$ (so the set of the orders of the elements of $G$ is stable by l.c.m.)

Proof. Put $m=|a|$ and $n=|b|$. We construct $c$ such that $|c|=m\vee n$. There are positive integers $m^{\prime }$ and $n^{\prime }$ such that $m\vee n=m^{\prime }{n}^{\prime }$, where $m^{\prime }\mid m,n^{\prime }\mid n$ and $m^{\prime }\wedge n^{\prime }=1$. To show the existence of $m^{\prime }$ and $n^{\prime }$, we write the decompositions of $m,n$ in prime factors

$$\begin{array}{llll}\hfill m& =p_1^{a_1}{p}_2^{a_2}\cdots p_k^{a_k},& \hfill & \\ \hfill n& =p_1^{b_1}{p}_2^{b_2}\cdots p_k^{b_k},(a_i\ge 0,b_i\ge 0).& \hfill & \end{array}$$

Then

$$\begin{array}{llll}\hfill m\vee n& =p_1^{c_1}{p}_2^{c_2}\cdots p_k^{c_k},\text{where }{c}_i=\max <!--\; FUNCTION\; APPLICATION\; -->(a_i,b_i).& \hfill & \end{array}$$

Put

$$\begin{array}{llll}\hfill m^{\prime }& =p_1^{d_1}{p}_2^{d_2}\cdots p_k^{d_k},& \hfill & \\ \hfill n^{\prime }& =p_1^{e_1}{p}_2^{e_2}\cdots p_k^{e_k},& \hfill & \end{array}$$

where

$$\begin{array}{lll}\hfill d_i=\{\begin{array}{cc}{a}_i\hfill & \text{if }{a}_i\ge b_i\hfill \\ 0\hfill & \text{if }{a}_i<b_i,\hfill \end{array}{e}_i=\{\begin{array}{cc}0\hfill & \text{if }{a}_i\ge b_i\hfill \\ b_i\hfill & \text{if }{a}_i<b_i.\hfill \end{array}& & \hfill \end{array}$$

Then $d_i\le a_i$ and $e_i\le b_i$ for all $i$, thus $m^{\prime }\mid m$ and $n^{\prime }\mid n$. Moreover $m^{\prime }$ and $n^{\prime }$ have no prime factor in common, so $m^{\prime }\wedge n^{\prime }=1$. Finally

$$m^{\prime }{n}^{\prime }=p_1^{d_1+e_1}{p}_2^{d_2+e_2}\cdots p_k^{d_k+e_k},$$

where $d_i+e_i=a_i=\max <!--\; FUNCTION\; APPLICATION\; -->(a_i,b_i)$ if $a_i\ge b_i$, and $d_i+e_i=b_i=\max <!--\; FUNCTION\; APPLICATION\; -->(a_i,b_i)$ if $a_i<b_i$. In both cases, $d_i+e_i=\max <!--\; FUNCTION\; APPLICATION\; -->(a_i,b_i)=c_i$, therefore $m^{\prime }{n}^{\prime }=m\vee n$.

Put $u=a^{m∕m^{\prime }}$ and $v=b^{n∕n^{\prime }}$. Then $|u|=m^{\prime }$ and $|v|=n^{\prime }$, where $m^{\prime }\wedge n^{\prime }=1$. Put $c=\mathit{uv}$. By Lemma 1, $|c|=|\mathit{uv}|=m^{\prime }{n}^{\prime }=m\vee n$. This proves the lemma. □

Lemma 3. Let $G$ be a finite abelian group. There exists some $a\in G$ such that $|a|=\omega (G)=l.c.m.\{|x|\mid x\in G\}$.

Proof. Let $G=\{x_1,x_2,\dots ,x_n\}$. By Lemma 2, we build an element $a$ of order $d=|x_1|\vee |x_2|\vee \cdots \vee |x_n|$ inductively. If $a_i\in G$ has order $d_i=|x_1|\vee |x_2|\vee \cdots \vee |x_i|$, where $k<n$, then there exists by Lemma 2 an element of order $d_{i+1}=d_i\vee |x_{i+1}|=|x_1|\vee |x_2|\vee \cdots \vee |x_i|\vee |x_{i+1}|$. □

Proof. (of Exercise 19 (b)) Let $(G,\cdot )$ be an abelian finite group. Assume for contradiction that $G$ is divisible. By definition, $n=|G|>1$. By Lemma 3, there is some element $a\in G$ such that $|a|=\omega (G)=l.c.m.\{|x|\mid x\in G\}$.

Since $G$ is not trivial, $e=\omega (G)>1$, so there is some prime $p$ such that $p\mid e$.

Moreover, since $G$ is divisible, there exists some element $x$ such that $a=x^p$. Then

$$a^{e∕p}={(x^p)}^{e∕p}=x^e=1.$$

This is impossible, because $|a|=e$ and $1\le e∕p<e$. This contradiction shows that no finite abelian group is divisible. □