Question

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

Short answer

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

Step-by-step solution

finite abelian p-group

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

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

Thus, $\mathrm{a}\notin \mathrm{pG}$.

pG≠G

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

$\begin{array}{rcl}\left|\mathrm{pb}\right|& =& \left|\mathrm{a}\right|\\ & =& {\mathrm{p}}^{\mathrm{m}}\end{array}$

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

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