Question

Question: Prove that the function $\mathit{f}$ in the proof of Theorem 7.19(1) is a bijection.

Short answer

It has been proved that $f$is a bijection.

Step-by-step solution

Step-By-Step SolutionStep 1: State f from the proof of Theorem 7.19(1)

Let G be an infinite cyclic group.

The function is defined as,

$f:G\to \mathrm{\mathbb{Z}}$

such that ,$f\left(a^k\right)=k$AND$k\in \mathrm{\mathbb{Z}}$

Show that the map is well-defined and surjective

Since has infinite order thus, all$a^k$

are distinct.

So, the power associated with any element of G is unique.

Thus, $f$is a well-defined map.

Now, for ,$k\in \mathrm{\mathbb{Z}},k=f\left(a^k\right)$

Thus, $f$is surjective.

Show that the map is injective

Let for $k,m\in \mathrm{\mathbb{Z}}$,$f\left(a^k\right)=f\left(a^m\right)$

Now ,$f\left(a^k\right)=k$and $f\left(a^m\right)=m$

thus $k=m$

Hence $f$is injective.

Conclusion

Since $f$is injective as well as surjective, therefore, $f$is a bijection.