Question

(a) If$\mathit{G}$is a finite abelian p-group, prove that$\mathit{p}\mathit{G}\mathbf{\ne }\mathit{G}$ .

Short answer

If G is a finite abelian p-group then $pG\ne G$.

Step-by-step solution

finite abelian p-group 

Let G be a finite abelian p-group. Then there exists an element $a\in G$ with maximal order as $\left|a\right|=p^m$ .

Since, a has maximal order the other elements in G will have order $p^n$where $n\le m$.

Thus, $a\notin pG$.

  

Assume that there is $b\in G$such that $pb=a$ then the order will be,

$\left|pb\right|=a$

localid="1657361911843" $=p^m$

Thus, $\left|pb\right|=p^m$implies $\left|b\right|=p^{m+1}$which contradicts the maximal order of a. Thus, $a\notin pG$.

Therefore, if G is a finite abelian p-group then $pG\ne G$.