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Statement: Every infinite cyclic group is isomorphic to additive group of integers. Proof: Let $G$ be infinite cyclic group with generator a. Claim: Two distinct powers of '$a$' cannot give same elements of $G$. For, Let. for $r,s$ in some positive integer, $a^ r=a ^s$ implies that $a^{r-s}=e$ implies that order of a is less than or equal to $r-s$ (finite).

But this contradicts that $G$ is infinite cyclic group generated by a. So I have a question here... How can we say that above step contradicts the given statement of having generator a? Also what has the claim to do to prove the theorem? Then the proof proceeds to show that a and $a^{-1}$ are generators. Claim: G has only two generators a and $a^{-1}$. For, if possible, let $a^m$ , where m is not equal to 1 &-1, is generator of G. Then for some k such that $(a^m)^k=a$.

My question here is how we get a for $(a^m)^k=a$?

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Without loss of generality assume $r > s$. Let $m = r - s$, so $a^m = e$. Every $n \in \mathbb{Z}$ can be written as $n = km + l$ for some $k \in \mathbb{Z}$ and some $l$ with $0 \le l < m$. Then:

$$\begin{align} a^n &= a^{km + l} \\ &= a^{mk}a^l \\ &= (a^m)^k a^l \\ &= e^k a^l \\ &= a^l \end{align} $$ So there are are only (at most) $m$ elements in $G$: $e = a^0, a^1, \dots, a^l, \dots, a^{m-1}$. ("At most" because we didn't assume that $m$ was the least $i$ such that $a^i = e$. The number of distinct elements actually divides $m$.)

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  • $\begingroup$ So order of a is less than or equal to m, right? $\endgroup$
    – Kavita
    Oct 10, 2015 at 4:21
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    $\begingroup$ Yes, if $a$ is a generator and has finite order. If $a^m = e$ then the order of $a$ actually divides $m$. (I added this to the answer.) $\endgroup$
    – BrianO
    Oct 10, 2015 at 4:24

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