Proving that any finite cyclic group of order $n$ is isomorphic to $$

Question

The proof goes on like this :

$Z$ denotes integers.

$a^m=e$ for some positive integer $m$. Let $n$ be the smallest positive integer such that $a^n=e$ if $s \epsilon Z$ and $s=nq+r$ for $0\le r < m$ then $$a^s=a^{nq+r}=a^r \ \ \ (*)$$.

Then after this statement, the book proves that all elements of the finite cyclic group is distinct and creates a homorphism between this group and $<Z_{n},+_{n}>$

What I couldn't get in the proof is what does $(*)$ tell us? I couldn't understand it and why is it needed in the proof? Also why do we work with positive integers, why do we let $n$ be the smallest integer instead of minimizing $|n|?$

$(*)$ is needed to show that every subgroup of a cyclic group is cyclic. For your question, see here. — Dietrich Burde, Mar 11 '17 at 21:32
You told us what $Z$ meant but you didn't tell us your definition of $Z_n$ and $+_n$. — Derek Holt, Mar 11 '17 at 22:27
There is a theorem saying that any two cyclic groups of the same order are isomorphic (cf. e.g., Nathan Jacobson, "Basic Algebra I", Sec 1.5, Th 1.2). — user5280911, Dec 23 '20 at 09:31

score 3 · Accepted Answer · edited Jul 23 '17 at 12:50

sea $G= \langle a \rangle$ con $\vert G \vert = n$ y $f: G \rightarrow \mathbb{Z}_n$. considera la función $a^n \mapsto n$. veamos que está bien definida: sean $h \neq k$ tales que $a^h = a^k$ entonces $a^h a^{-k} = e = a^{h-k}$ pero $h-k < n$ el orden de $a$ lo cual es una contradicción, por tanto, $f$ está bien definida y además es INYECTIVA. veamos que $f$ es un morfismo: $f(a^h a^k) = f(a^{h+k})= h + k = f( a^h)+ f( a^k )$ y por tanto es MORFISMO. finalmente veamos que $f$ es suprayectiva: sea $x \in \mathbb{Z}_n$ entonces $0 \leq x < n$ y entonces existe $a^x \in G$ tal que $f( a^x) = x$. y entonces $f$ es un ISOMORFISMO.

Let $G = \langle a\rangle$ with $\lvert G\rvert = n$ and $f \colon G \to \mathbb{Z}_n$. Consider the function $a^n \mapsto n$. We shall see it is well-defined: Let $h \neq k$ such that $a^h = a^k$, therefore $a^ha^{-k} = e = a^{h-k}$, but $h - k < n$, the order of $a$, so we have a contradiction. Thus $f$ is well-defined, and also injective. Let's show that $f$ is a homomorphism: $f(a^ha^k) = f(a^{h+k}) = h+k = f(a^h) + f(a^k)$, so $f$ is a homomorphism. Finally, we see that $f$ is surjective: let $x \in \mathbb{Z}_n$, therefore $0 \leqslant x < n$ and hence there is $a^x \in G$ such that $f(a^x) = x$. Thus $f$ is an isomorphism.

In this case, can $x$ be taken less than the order of $a$? – wavilson ferreira Dec 09 '18 at 01:53 — wavilson ferreira, Dec 09 '18 at 01:53

score 0 · Answer 2 · 2017-03-11T21:44:50.370

I am guessing the homomorphism your book constructed was $a^k\to k(\text{ mod } m)$, in which case (*) shows that the function is injective.

After all, if $a^s\neq a^r$, then we would have two distinct elements going to $r$ under your homomorphism.

And if we only want to minimize the absolute value of $n$, then we loose the uniqueness of $r$ and a lot of nice properties which you will learn about when studying Euclidean Domains.

Proving that any finite cyclic group of order $n$ is isomorphic to $$

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