Let $G = \langle a\rangle$ a cyclic group of order m. Prove that G is isomorphic with $Z_m$.

Question

Let $f$: $Z_m \rightarrow G$, $f(n)$ = $a^n$. Show that f is a function and an homomorphism.

Doing a review on the whole semester, and a little refresh on this topic would be great.

By definition I know $G$ has the same order of $Z_m$, but not sure how to attack this problem with ease.

I also know that cyclic groups of the same order are isomorphic.

(I cannot use the fact that cyclic groups of the same order are isomorphic, that's why I ask for another approach, because I can't use theorems or lemmas that have not been discussed in class.)

Answer 1

First, show that $f$ is a function:

Let $f(q) = a^q$ and $f(q+km) = a^{q+km}$

$a^{q+km}=a^qa^{km}=a^qe=a^q$

Hence, $f$ is correctly defined.

Next, prove that $f$ is a homomorphism:

$f(qs) = a^{qs} = a^qa^s = f(q)f(s)$

Then, prove that $f$ is injective:

Let $f(q) = a^q$ and $f(s) = a^s$. If $a^q = a^s$, then $q=s $ mod $ m$, then $q=s$.

Since $|Z_m| = |G|$ and $|Z_m| < \infty$, $f$ is an isomorphism.