Question

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

Short answer

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

Step-by-step solution

f is injective

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

The kernel of f is $kerf=\{g\in G:f(g)=ng=0\}$ . Let $g\in kerf$has order localid="1657345446245" $p^k$then $\left(n,p^k\right)=1$ implies there exists element $x,y\in \square$such that $nx+y{p}^k=1$.

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

localid="1657345467831" $g=1·g$

$=\left(nx+y{p}^k\right)g$

localid="1657302699494" $=gnx+gy{p}^k$

$=0$

Therefore, f is injective.

f is a homomorphism

Let $a,b\in G$ then,

$f(a+b)=n(a+b)$

$=na+nb$

$=f\left(a\right)+f\left(b\right)$

Thus, f is a homomorphism.

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