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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.)

user2371916
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  • If you know that cyclic groups of the same order are isomorphic, then you would just like to show that $Z_m$ is cyclic and of order $m$, yes? Try doing that. – Tai Jul 26 '17 at 03:46
  • Have you tried using the First Isomorphism Theory with the funciton $f$ as your morfism? – BB3C Jul 26 '17 at 03:49

1 Answers1

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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.

  • I was about to comment on showing that f was a function.

    The fact that f is correctly defined + the fact is a homorphism is enough to call it an isomorphism because is also onto, right? Just double checking for future references.

     – user2371916 Jul 26 '17 at 04:11
  • First it is necessary to show that $f$ defined as $\mathbb{Z} \rightarrow G: f(n) = a^n$ actually satisfies the definition of a function. In order to show that it is an isomorphism, it is necessary to show that it satisfies $f(ab) = f(a)f(b)$ and that $|Z_m| = |G|$. –  Jul 26 '17 at 04:14
  • Well you also have to show that $f$ is injective or surjective. The trivial homomorphism of sending everything to the neutral element is well defined and an homomorphism but obviously not an isomorphism. – Verdruss Jul 26 '17 at 08:12
  • @ChristopherDeJesus As an extra hint towards showing injectivity/surjectivity, recall that for any group element $g\in G$ the homomorphism $\epsilon_g:\mathbb Z \to G$ defined by $\epsilon_g(a)=g^a$ is surjective if and only if $g$ generates $G$. – Andrew Tawfeek Jul 26 '17 at 08:35
  • @Verdruss Thanks, I have edited the answer –  Jul 26 '17 at 13:06