Problem 5.2.5 (Exponent of an abelian group)

Proof. By hypothesis

where $n_{i+1}\mid n_i$ for $1\le i<t$. Moreover $Z_{n_i}=⟨x_i⟩$, where $|x_i|=n_i$ for all $i$.

Pick $a=(a_1,a_2,\dots ,a_t)\in \tilde{G}$. Since $a_i\in Z_{n_i}=⟨x_i⟩$, then $a_i=x_i^{k_i}$ for some integer $k_i$. Moreover $n_1=q_i{n}_i$ for some integer $q_i$, thus $a_i^{n_1}={(x_i^{n_i})}^{k_i{q}_i}=1$, so

Since $G\simeq \tilde{G}$, for all $x\in G,$

Suppose that $G$ contains an element $x$ of order $m$. Since $x^{n_1}=1$, we obtain

Conversely, assume that $m\mid n_1$, so that $n_1=\mathit{𝑞𝑚}$ for some positive integer $q$. Put

Then, for all integers $k$,

The equivalence $a^k=1⟺m\mid k$, true for all integers $k$, shows that $|a|=m$, so $\tilde{G}$ contains an element $a$ of order $m$.

Since $G\simeq \tilde{G}$, $G$ contains an element $x$ of order $m$.

In conclusion, $G$ contains an element of order $m$ if and only if $m\mid n_1$.

By (1), $x^{n_1}=1$ for all $x\in G$. Suppose now that a positive integer $n$ satisfies $x^n=1$ for all $x\in G$. By the first part, there is some element $x_1$ of order $n_1$ in $G$. Then $x_1^n=1$, therefore $n_1\mid n$, where $n_1>0,n>0$, thus $n_1\le n$.

This proves that $n_1$ is the smallest positive integer $n$ such that $x^n=1$ for all $x\in G$. So $n_1$ is the exponent of $G$ (see definition p. 165). □