Question

If G is an abelian p-group and $\left(n,p\right)\mathbf{=}\mathbf{1}$ prove that the map $\mathbf{f}\mathbf{:}\mathbf{G}\mathbf{\to }\mathbf{G}$ given by $\mathbf{f}\left(a\right)\mathbf{=}\mathbf{na}$ is an isomorphism.

Short answer

The map $\mathrm{f}:\mathrm{G}\to \mathrm{G}$ defined by$\mathrm{f}\left(\mathrm{a}\right)=\mathrm{na}$ is an isomorphism.

Step-by-step solution

f is injective

Let G be an abelian p-group and $\left(\mathrm{n},\mathrm{p}\right)=1$. Define the map $\mathrm{f}:\mathrm{G}\to \mathrm{G}$given by $\mathrm{f}\left(\mathrm{a}\right)=\mathrm{na}$.

The kernel of f is $\ker \mathrm{f}=\left\{\mathrm{g}\in \mathrm{G}:\mathrm{f}\left(\mathrm{g}\right)=\mathrm{ng}=0\right\}$. Let $\mathrm{g}\in \ker \mathrm{f}$has order ${\mathrm{p}}^{\mathrm{k}}$then $\left(\mathrm{n},{\mathrm{p}}^{\mathrm{k}}\right)=1$implies there exists element $\mathrm{x},\mathrm{y}\in \cancel{\mathrm{c}}$such that $\mathrm{nx}+{\mathrm{yp}}^{\mathrm{k}}=1$.

Since, $\mathrm{ng}=0$ implies $\mathrm{xng}=0$ and ${\mathrm{yp}}^{\mathrm{k}}\mathrm{g}=0$ then for any element $\mathrm{g}\in \mathrm{G}$ is,

$\begin{array}{rcl}\mathrm{g}& =& 1.\mathrm{g}\\ & =& \left(\mathrm{nx}+{\mathrm{yp}}^{\mathrm{k}}\right)\mathrm{g}\\ & =& \mathrm{gnx}+{\mathrm{gyp}}^{\mathrm{k}}\\ & =& 0\end{array}$

Therefore, f is injective.

f is a homomorphism

Let $\left(\mathrm{a},\mathrm{b}\right)\in \mathrm{G}$ then,

$\begin{array}{rcl}\mathrm{f}\left(\mathrm{a}+\mathrm{b}\right)& =& \mathrm{n}\left(\mathrm{a}+\mathrm{b}\right)\\ & =& \mathrm{na}+\mathrm{nb}\\ & =& \mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}\left(\mathrm{b}\right)\end{array}$

Thus, f is a homomorphism.

Therefore, the map $\mathrm{f}:\mathrm{G}\to \mathrm{G}$ defined by $\mathrm{f}\left(\mathrm{a}\right)=\mathrm{na}$ is an isomorphism.