Question

Assume that $\mathit{a}$and$\mathbf{b}$ are both generators of the cyclic group$\mathbf{G}$ so that $\mathbf{G}\mathbf{=}\mathbf{⟨}\mathbf{a}\mathbf{⟩}$ and$\mathit{G}\mathbf{=}\mathbf{⟨}\mathit{b}\mathbf{⟩}$ . Prove that the function$\mathbf{f}\mathbf{:}\mathbf{G}\mathbf{\to }\mathbf{G}$ given by$\mathit{f}\mathbf{\left(}{\mathbf{a}}^{\mathbf{j}}\mathbf{\right)}\mathbf{=}{\mathit{b}}^{\mathbf{j}}$ is an automorphism of $\mathit{G}$.

Short answer

It has been proved that function$f:G\to G$ given by$f\left(a^j\right)=b^j$ is an automorphism of$G$ .

Step-by-step solution

Given that

Given that $\mathrm{a}$and $\mathrm{b}$are both generators of the cyclic group $\mathrm{G}$such that $\mathrm{G}=⟨\mathrm{a}⟩$ and $\mathrm{G}=⟨\mathrm{b}⟩$.

Also function role="math" localid="1654325448250" $\mathrm{f}:\mathrm{G}\to \mathrm{G}$is given by $\mathrm{f}\left({\mathrm{a}}^{\mathrm{j}}\right)={\mathrm{b}}^{\mathrm{j}}$.

Show automorphism when G is infinite

If G is infinite then any element of G say g is uniquely represented as $\mathrm{g}={\mathrm{a}}^{\mathrm{m}}$for some $\mathrm{m}\in \mathrm{\mathbb{Z}}$.

Then,$f\left(a^m\right)=b^m$ is well defined.

The homomorphism property is easily seen by the law of exponents.

Since b is also a generator, it is clear that f is an isomorphism, hence an automorphism.

Show automorphism when G is finite

If G is a finite group of order n then$\mathrm{g}={\mathrm{a}}^{\mathrm{m}}$representation is not unique.

Now, if a is a generator of G, then there is an isomorphism ${\mathrm{\phi }}_{\mathrm{a}}:{\mathrm{\mathbb{Z}}}_{\mathrm{n}}\to \mathrm{G}$with ${\mathrm{\phi }}_{\mathrm{a}}\left(1\right)=\mathrm{a}$.

Define $\mathrm{f}={\mathrm{\phi }}_{\mathrm{b}}\circ {{\mathrm{\phi }}^{-1}}_{\mathrm{a}}:\mathrm{G}\to \mathrm{G}$

This implies fis an isomorphism and $\mathrm{f}\left(\mathrm{a}\right)={\mathrm{\phi }}_{\mathrm{b}}\left(1\right)=\mathrm{b}$

Conclusion

Hence, f is an automorphism.