Question

Let $\mathit{G}$be an abelian group of order$\mathit{n}$ and let$\mathit{K}$ be a positive integer. If $\mathbf{(}\mathbf{k}\mathbf{,}\mathbf{n}\mathbf{)}\mathbf{=}\mathbf{1}$, prove that the functionrole="math" localid="1654351034332" $\mathit{f}\mathbf{:}\mathit{G}\mathbf{\to }\mathit{G}$ given by$\mathit{f}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{=}{\mathit{a}}^{\mathbf{k}}$ is an isomorphism.

Short answer

It is proved that$f$ is isomorphism.

Step-by-step solution

Determine function is an isomorphism

Consider the given functions, $G$is the abelian group of order $n$and $k$is the positive integer, so that $(n,k)=1$.

Firstly observe that,

$g,h\in G$then,

$f\left(gh\right)={\left(gh\right)}^k=g^k{h}^k=f\left(g\right)f\left(h\right)$

Here, second equality holds as$G$ is abelian. Therefore,$f$ is a group of isomorphism.

Determine further simplification of function is an isomorphism

Now let’s consider, $g\in G$so that $f\left(g\right)=e$. That means,

$g^k=e$

That implies that the order of $g$divides$k$.

By Lagrange’s theorem it is clear that the order of $g$divides the order of $G$.

Therefore, it is clear that, $g$is the common divisor for $k$and n. But as$(n,k)=1$then the order of $g=1$. Thus$g=e$.

Now, $g_1,g_2\in G$then,

role="math" localid="1654351546623" $f\left(g_1\right)=f\left(g_2\right)$

That is,

${g_1}^k={g_2}^k$

Implies that,

${\left(g_1{g_2}^{-1}\right)}^k=e$

Form above all equations it is clear that, $f$is injective.

Determine whether further simplification of function is an isomorphism

As$G$is a finite set and$f:G\to G$is the injective function and also$f$in surjective.

As,$(n,k)=1$there exist two integers $x$and $y$, so that,

$nx+ky=1$

Thus it follows ,$h\in G$

$h=h^{nx+ky}=h^{nx}{h}^{ky}=f\left(h^y\right)$

The equality follows corollary 8.6, thus $h\in G$,

$f\left(h^y\right)=h$

Here$y$ is an integer, so that $(n,k)=1$ for some integer $x$.

Therefore,$f$ is surjective.

And hence$f$ is isomorphism.