Exercise 4.15

Proof. Let $n=|G|$. From Lagrange’s theorem, $a^n=1$ for all $a\in G$, so the polynomial $x^n-1\in K[x]$ has exactly $n$ distinct roots in $G$, and so

If $d\mid n$, the polynomial $x^d-1\in K[x]$ has exactly $d$ roots in $K$ otherwise $x^n-1=(x^d-1)g(x),g(x)\in K[x]$, and $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)=n-d$ has at most $n-d$ roots, so $x^n-1$ would have less than $n$ roots in $K$. As $x_0^d=1\Rightarrow x_0^n=1$, all these roots are in $G$ : $x^d-1$ has $d$ roots in $G$.

Let $\psi (d)$ the number of elements in $G$ with order $d$ ( $\psi (d)=0$ if $d\nmid n$). Then ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{c\mid d}\psi (c)=d$. Applying the Möbius inversion theorem, $\psi (d)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{c\mid d}\mu (c)d∕c=\varphi (d)$ (Prop. 2.2.5), in particular, $\psi (n)=\varphi (n)\ge 1$. This proves the existence of an element of order $n$ in $G$, so $G$ is cyclic.

(Variation : $\psi (d)=0$ if there exists no element of order $d$, and $\psi (d)=\varphi (d)$ otherwise (see Ex.4.13). So $\psi (d)\le \varphi (d)$ for all $d\mid n$. As ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{d\mid n}\psi (d)={\sum }_{d\mid n}\varphi (d)=n$, $\psi (d)=\varphi (d)$ for all $d\mid n$. So there exists in $G$ an element of order $n$, and $G$ is cyclic.) □

Proof. Let $n=|G|=p_1^{a_1}\cdots p_k^{a_k}$. From Lagrange’s theorem, $y^n=1$ for all $y\in G$.

$p(x)=x^{n∕p_1}-1\in K[x]$ has at most $n∕p_1<n$ roots in $K^{\text{∗}}$, a fortiori in $G$, so there exists $a\in G$ such that $a^{n∕p_1}\ne 1$.

Let $c_1=a^{n∕p_1^{a_1}}=a^{p_2^{a_2}\cdots p_k^{a_k}}$. Then $c_1^{p_1^{a_1}}=1$ and $c_1^{p_1^{a_1-1}}=a^{n∕p_1}\ne 1$, so $|c_1|=p_1^{a_1}$.

Similarly, there exist $c_2,\dots ,c_k$ with respective orders $|c_i|=p_i^{a_i}$.

From exercise 4.14, we obtain by induction that $c=c_1\cdots c_k$ has order $p_1^{a_1}\cdots p_k^{a_k}=n$, so $G$ is cyclic. □